RS Aggarwal Solutions for Class 11 Chapter 10 Binomial Theorem

Question 1. Using binomial theorem, expand each of the following: (1 – 2x)⁵
Answer: To expand \( (1 - 2x)^5 \), we apply the binomial expansion formula. Substituting into \( (a+b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r \), we get:
\[ \binom{5}{0}(1)^5 + \binom{5}{1}(1)^4(-2x) + \binom{5}{2}(1)^3(-2x)^2 + \binom{5}{3}(1)^2(-2x)^3 + \binom{5}{4}(1)(-2x)^4 + \binom{5}{5}(-2x)^5 \]
Simplifying each term:
\[ 1 - 5(2x) + 10(4x^2) - 10(8x^3) + 5(16x^4) - 1(32x^5) \]
\[ = 1 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5 \]
Rearranging in descending powers:
\[ -32x^5 + 80x^4 - 80x^3 + 40x^2 - 10x + 1 \]
In simple words: To expand a binomial raised to a power, multiply out using the binomial formula where each term involves a binomial coefficient, decreasing powers of the first part, and increasing powers of the second part.

Exam Tip: Check the pattern of signs - when the second term is negative, signs alternate. Always arrange the final answer in descending order of powers for clarity.

Question 2. Using binomial theorem, expand each of the following: (2x – 3)⁶
Answer: To expand \( (2x - 3)^6 \), we use the binomial expansion formula with \( a = 2x \) and \( b = -3 \):
\[ \binom{6}{0}(2x)^6 + \binom{6}{1}(2x)^5(-3) + \binom{6}{2}(2x)^4(-3)^2 + \binom{6}{3}(2x)^3(-3)^3 + \binom{6}{4}(2x)^2(-3)^4 + \binom{6}{5}(2x)(-3)^5 + \binom{6}{6}(-3)^6 \]
Computing each term:
\[ (1)(64x^6) - (6)(32x^5)(3) + (15)(16x^4)(9) - (20)(8x^3)(27) + (15)(4x^2)(81) - (6)(2x)(243) + (1)(729) \]
\[ = 64x^6 - 576x^5 + 2160x^4 - 4320x^3 + 4860x^2 - 2916x + 729 \]
In simple words: Each term is found by multiplying binomial coefficients with decreasing powers of 2x and increasing powers of -3. Alternating signs appear because the second quantity is negative.

Exam Tip: When coefficients (like 2) appear in the original binomial, raise them to their respective powers in each term - missing this step is a common mistake.

Question 3. Using binomial theorem, expand each of the following: (3x + 2y)⁵
Answer: To expand \( (3x + 2y)^5 \), we apply the binomial formula with \( a = 3x \) and \( b = 2y \):
\[ \binom{5}{0}(3x)^5 + \binom{5}{1}(3x)^4(2y) + \binom{5}{2}(3x)^3(2y)^2 + \binom{5}{3}(3x)^2(2y)^3 + \binom{5}{4}(3x)(2y)^4 + \binom{5}{5}(2y)^5 \]
Computing:
\[ (1)(243x^5) + (5)(81x^4)(2y) + (10)(27x^3)(4y^2) + (10)(9x^2)(8y^3) + (5)(3x)(16y^4) + (1)(32y^5) \]
\[ = 243x^5 + 810x^4y + 1080x^3y^2 + 720x^2y^3 + 240xy^4 + 32y^5 \]
In simple words: Each term combines a binomial coefficient with powers of 3x and 2y. The power of x decreases while the power of y increases, and all coefficients are positive since both original terms are positive.

Exam Tip: When both terms in the binomial are positive, all terms in the expansion remain positive - this is a quick check on your work.

Question 4. Using binomial theorem, expand each of the following: (2x – 3y)⁴
Answer: To expand \( (2x - 3y)^4 \), we use the binomial expansion with \( a = 2x \) and \( b = -3y \):
\[ \binom{4}{0}(2x)^4 + \binom{4}{1}(2x)^3(-3y) + \binom{4}{2}(2x)^2(-3y)^2 + \binom{4}{3}(2x)(-3y)^3 + \binom{4}{4}(-3y)^4 \]
Evaluating each term:
\[ (1)(16x^4) - (4)(8x^3)(3y) + (6)(4x^2)(9y^2) - (4)(2x)(27y^3) + (1)(81y^4) \]
\[ = 16x^4 - 96x^3y + 216x^2y^2 - 216xy^3 + 81y^4 \]
In simple words: The expansion alternates between positive and negative terms because the second part is negative. The powers of x go down while the powers of y go up with each successive term.

Exam Tip: With even powers (like 4 here), the first and last terms are always positive - use this to verify your signs are correct.

Question 5. Using binomial theorem, expand each of the following: \( \left(\frac{2x}{3} - \frac{3}{2x}\right)^6 \)
Answer: To expand \( \left(\frac{2x}{3} - \frac{3}{2x}\right)^6 \), we apply the binomial formula with \( a = \frac{2x}{3} \) and \( b = -\frac{3}{2x} \):
\[ \binom{6}{0}\left(\frac{2x}{3}\right)^6 + \binom{6}{1}\left(\frac{2x}{3}\right)^5\left(-\frac{3}{2x}\right) + \binom{6}{2}\left(\frac{2x}{3}\right)^4\left(-\frac{3}{2x}\right)^2 + \binom{6}{3}\left(\frac{2x}{3}\right)^3\left(-\frac{3}{2x}\right)^3 + \binom{6}{4}\left(\frac{2x}{3}\right)^2\left(-\frac{3}{2x}\right)^4 + \binom{6}{5}\left(\frac{2x}{3}\right)\left(-\frac{3}{2x}\right)^5 + \binom{6}{6}\left(-\frac{3}{2x}\right)^6 \]
Simplifying:
\[ 1 \cdot \frac{64x^6}{729} - 6 \cdot \frac{32x^5}{243} \cdot \frac{3}{2x} + 15 \cdot \frac{16x^4}{81} \cdot \frac{9}{4x^2} - 20 \cdot \frac{8x^3}{27} \cdot \frac{27}{8x^3} + 15 \cdot \frac{4x^2}{9} \cdot \frac{81}{16x^4} - 6 \cdot \frac{2x}{3} \cdot \frac{243}{32x^5} + 1 \cdot \frac{729}{64x^6} \]
\[ = \frac{64x^6}{729} - \frac{32x^4}{27} + \frac{20x^2}{3} - 20 + \frac{135}{4x^2} - \frac{243}{8x^4} + \frac{729}{64x^6} \]
In simple words: Each term involves powers of fractions, so we raise both numerator and denominator to their respective powers. The powers of x gradually shift from positive to negative across the expansion.

Exam Tip: When fractions appear, simplify each term completely before combining - keeping track of powers in both numerator and denominator prevents sign and calculation errors.

Question 6. Using binomial theorem, expand each of the following: \( \left(x^2 - \frac{3}{x}\right)^7 \)
Answer: To expand \( \left(x^2 - \frac{3}{x}\right)^7 \), we use the binomial expansion with \( a = x^2 \) and \( b = -\frac{3}{x} \). Applying the formula and computing each binomial coefficient:
\[ \binom{7}{0}(x^2)^7 + \binom{7}{1}(x^2)^6\left(-\frac{3}{x}\right) + \binom{7}{2}(x^2)^5\left(-\frac{3}{x}\right)^2 + \binom{7}{3}(x^2)^4\left(-\frac{3}{x}\right)^3 + \binom{7}{4}(x^2)^3\left(-\frac{3}{x}\right)^4 + \binom{7}{5}(x^2)^2\left(-\frac{3}{x}\right)^5 + \binom{7}{6}(x^2)\left(-\frac{3}{x}\right)^6 + \binom{7}{7}\left(-\frac{3}{x}\right)^7 \]
Computing and simplifying:
\[ x^{14} - 7x^{12}\left(\frac{3}{x}\right) + 21x^{10}\left(\frac{9}{x^2}\right) - 35x^8\left(\frac{27}{x^3}\right) + 35x^6\left(\frac{81}{x^4}\right) - 21x^4\left(\frac{243}{x^5}\right) + 7x^2\left(\frac{729}{x^6}\right) - \frac{2187}{x^7} \]
\[ = x^{14} - 3x^{13} + \frac{27}{7}x^{12} - \frac{135}{49}x^{11} + \frac{405}{343}x^{10} - \frac{729}{2401}x^9 + \frac{729}{16807}x^8 - \frac{2187}{823543}x^7 \]
In simple words: For each term, the exponent of x decreases by combining positive powers from \( x^2 \) with negative powers from \( \frac{3}{x} \). The coefficients alternate in sign.

Exam Tip: Track the total power of x in each term by adding exponents (positive from \( x^2 \) and negative from \( x^{-1} \)) - this helps verify that powers form a complete sequence.

Question 7. Using binomial theorem, expand each of the following: \( \left(x^2 - \frac{3x}{7}\right)^7 \)
Answer: To expand \( \left(x^2 - \frac{3x}{7}\right)^7 \), we apply the binomial formula with \( a = x^2 \) and \( b = -\frac{3x}{7} \). Working through each term of the expansion:
\[ \binom{7}{0}(x^2)^7 + \binom{7}{1}(x^2)^6\left(-\frac{3x}{7}\right) + \binom{7}{2}(x^2)^5\left(-\frac{3x}{7}\right)^2 + \cdots + \binom{7}{7}\left(-\frac{3x}{7}\right)^7 \]
After computing all terms and simplifying the coefficients:
\[ x^{14} - 3x^{13} + \frac{27}{7}x^{12} - \frac{135}{49}x^{11} + \frac{405}{343}x^{10} - \frac{729}{2401}x^9 + \frac{729}{16807}x^8 - \frac{2187}{823543}x^7 \]
In simple words: Each term combines a binomial coefficient with decreasing powers of \( x^2 \) and increasing powers of \( -\frac{3x}{7} \). Powers of x decrease by 1 in each term while fractions become more complex.

Exam Tip: When expanding binomials with fractional coefficients, compute the numerical coefficients separately to avoid arithmetic mistakes in the final answer.

Question 8. Using binomial theorem, expand each of the following: \( \left(x - \frac{1}{y}\right)^5 \)
Answer: To expand \( \left(x - \frac{1}{y}\right)^5 \), we use the binomial formula with \( a = x \) and \( b = -\frac{1}{y} \):
\[ \binom{5}{0}(x)^5 + \binom{5}{1}(x)^4\left(-\frac{1}{y}\right) + \binom{5}{2}(x)^3\left(-\frac{1}{y}\right)^2 + \binom{5}{3}(x)^2\left(-\frac{1}{y}\right)^3 + \binom{5}{4}(x)\left(-\frac{1}{y}\right)^4 + \binom{5}{5}\left(-\frac{1}{y}\right)^5 \]
Evaluating each term:
\[ x^5 - 5 \cdot \frac{x^4}{y} + 10 \cdot \frac{x^3}{y^2} - 10 \cdot \frac{x^2}{y^3} + 5 \cdot \frac{x}{y^4} - \frac{1}{y^5} \]
\[ = x^5 - \frac{5x^4}{y} + \frac{10x^3}{y^2} - \frac{10x^2}{y^3} + \frac{5x}{y^4} - \frac{1}{y^5} \]
In simple words: Powers of x decrease from 5 to 0 while powers of y increase from 0 to 5 in the denominators. Signs alternate because the second term is negative.

Exam Tip: Check that the sum of the exponents of x and y in each term equals 5 - this verifies the structure of your expansion is correct.

Question 9. Using binomial theorem, expand each of the following: \( \left(\sqrt{x} + \sqrt{y}\right)^8 \)
Answer: To expand \( \left(\sqrt{x} + \sqrt{y}\right)^8 \), we first rewrite the expression as \( \left(x^{1/2} + y^{1/2}\right)^8 \) and apply the binomial formula:
\[ \binom{8}{0}(x^{1/2})^8 + \binom{8}{1}(x^{1/2})^7(y^{1/2}) + \binom{8}{2}(x^{1/2})^6(y^{1/2})^2 + \cdots + \binom{8}{8}(y^{1/2})^8 \]
Computing each term:
\[ x^4 + 8(x^{7/2})(y^{1/2}) + 28(x^3)(y) + 56(x^{5/2})(y^{3/2}) + 70(x^2)(y^2) + 56(x^{3/2})(y^{5/2}) + 28(x)(y^3) + 8(x^{1/2})(y^{7/2}) + y^4 \]
In simple words: Rewrite square roots as fractional exponents, then apply the binomial expansion. The exponents of x decrease by 1/2 with each term while the exponents of y increase by 1/2.

Exam Tip: Convert radicals to fractional exponents at the start - this prevents errors when managing powers throughout the expansion.

Question 10. Using binomial theorem, expand each of the following: \( \left(\sqrt[3]{x} - \sqrt[3]{y}\right)^6 \)
Answer: To expand \( \left(\sqrt[3]{x} - \sqrt[3]{y}\right)^6 \), we rewrite as \( \left(x^{1/3} - y^{1/3}\right)^6 \) and apply the binomial formula with \( a = x^{1/3} \) and \( b = -y^{1/3} \):
\[ \binom{6}{0}(x^{1/3})^6 + \binom{6}{1}(x^{1/3})^5(-y^{1/3}) + \binom{6}{2}(x^{1/3})^4(-y^{1/3})^2 + \binom{6}{3}(x^{1/3})^3(-y^{1/3})^3 + \binom{6}{4}(x^{1/3})^2(-y^{1/3})^4 + \binom{6}{5}(x^{1/3})(-y^{1/3})^5 + \binom{6}{6}(-y^{1/3})^6 \]
Computing:
\[ x^2 - 6(x^{5/3})(y^{1/3}) + 15(x^{4/3})(y^{2/3}) - 20(x)(y) + 15(x^{2/3})(y^{4/3}) - 6(x^{1/3})(y^{5/3}) + y^2 \]
In simple words: Cube roots are written as fractional exponents with denominator 3. The binomial expansion proceeds normally, with exponents decreasing by 1/3 for x and increasing by 1/3 for y in each term.

Exam Tip: Verify that fractional exponents are computed correctly by checking that each exponent sum equals 2 (the power of the original binomial divided by 3).

Question 11. Using binomial theorem, expand each of the following: (1 + 2x - 3x²)⁴
Answer: To expand \( (1 + 2x - 3x^2)^4 \), we group the first two terms and let \( a = 1 + 2x \) and \( b = -3x^2 \). The expression becomes \( (a + b)^4 \). After applying the binomial formula:
\[ 1 \cdot (1 + 2x)^4 - 4(1 + 2x)^3(3x^2) + 6(1 + 2x)^2(9x^4) - 4(1 + 2x)(27x^6) + 1(81x^8) \]
We must expand each power of \( (1 + 2x) \) separately using the binomial theorem:
\( (1 + 2x)^4 = 1 + 8x + 24x^2 + 32x^3 + 16x^4 \)
\( (1 + 2x)^3 = 1 + 6x + 12x^2 + 8x^3 \)
\( (1 + 2x)^2 = 1 + 4x + 4x^2 \)
Substituting and collecting like terms yields:
\[ 81x^8 - 216x^7 + 108x^6 + 120x^5 - 74x^4 - 40x^3 + 12x^2 + 8x + 1 \]
In simple words: When a binomial has three terms, group two of them as one unit. Apply the binomial formula once, then expand the grouped powers separately and combine all terms by collecting powers of x.

Exam Tip: Be systematic: expand each intermediate power completely before substituting back - this minimizes algebra mistakes when combining the final result.

Question 12. Using binomial theorem, expand each of the following: \( \left(1 + \frac{x}{2} - \frac{2}{x}\right)^4 \), where \( x \neq 0 \)
Answer: To expand \( \left(1 + \frac{x}{2} - \frac{2}{x}\right)^4 \), we set \( a = 1 + \frac{x}{2} \) and \( b = -\frac{2}{x} \). Applying the binomial formula:
\[ \left(1 + \frac{x}{2}\right)^4 - 4\left(1 + \frac{x}{2}\right)^3\left(\frac{2}{x}\right) + 6\left(1 + \frac{x}{2}\right)^2\left(\frac{4}{x^2}\right) - 4\left(1 + \frac{x}{2}\right)\left(\frac{8}{x^3}\right) + \frac{16}{x^4} \]
Expanding each power of \( \left(1 + \frac{x}{2}\right) \):
\( \left(1 + \frac{x}{2}\right)^4 = 1 + 2x + \frac{3}{2}x^2 + \frac{1}{2}x^3 + \frac{1}{16}x^4 \)
\( \left(1 + \frac{x}{2}\right)^3 = 1 + \frac{3}{2}x + \frac{3}{4}x^2 + \frac{1}{8}x^3 \)
\( \left(1 + \frac{x}{2}\right)^2 = 1 + x + \frac{1}{4}x^2 \)
Substituting and collecting like terms:
\[ \frac{1}{16}x^4 + \frac{1}{2}x^3 + \frac{3}{2}x^2 + 2x + 1 \]
In simple words: Group the first two terms as a unit, then apply binomial expansion twice - once to separate the three-term group, and again for each power of the grouped part. Carefully combine all resulting terms.

Exam Tip: Check your answer by substituting a simple value (like \( x = 2 \)) into both the original and expanded forms - they should yield the same result.

Question 12. Using binomial theorem, expand each of the following: (3x² - 2ax + 3a²)³
Answer: We need to find the expansion of (3x² - 2ax + 3a²)³
Formula: (a+b)ⁿ = ⁿC₀aⁿ + ⁿC₁aⁿ⁻¹b + ⁿC₂aⁿ⁻²b² + … + ⁿCₙbⁿ
Let p = 3x² - 2ax, so the expression becomes (p + 3a²)³
Using the binomial formula with a = 3x² and b = -2ax:
For (a+b)³: a³ + b³ + 3a²b + 3ab²
(3x² - 2ax)³ = 27x⁶ - 8a³x³ - 54ax⁵ + 36a²x⁴
For (a+b)²: a² + 2ab + b²
(3x² - 2ax)² = 9x⁴ - 12x³a + 4a²x²
Substituting both results back into the binomial expansion:
\( = [1(27x^6 - 8a^3x^3 - 54ax^5 + 36a^2x^4)] + [3(9x^4 - 12x^3a + 4a^2x^2)(3a^2)] + [3(3x^2 - 2ax)(9a^4)] + [1(27a^6)] \)
\( = 27x^6 - 54ax^5 + 117a^2x^4 - 116x^3a^3 + 117a^4x^2 - 54a^5x + 27a^6 \)

Exam Tip: When expanding trinomials, substitute to create a binomial form first. Group terms strategically and carefully track all coefficient multiplications through each step.

Question 13. Evaluate: \( \left(\sqrt{2}+1\right)^6+\left(\sqrt{2}-1\right)^6 \)
Answer: We need to find the value of \( \left(\sqrt{2}+1\right)^6+\left(\sqrt{2}-1\right)^6 \)
Formula: (a+1)⁶ = ⁶C₀a⁶ + ⁶C₁a⁵ + ⁶C₂a⁴ + ⁶C₃a³ + ⁶C₄a² + ⁶C₅a + ⁶C₆
(a-1)⁶ = ⁶C₀a⁶ - ⁶C₁a⁵ + ⁶C₂a⁴ - ⁶C₃a³ + ⁶C₄a² - ⁶C₅a + ⁶C₆
When we add these two expansions, the odd-powered terms cancel:
(a+1)⁶ + (a-1)⁶ = 2[⁶C₀a⁶ + ⁶C₂a⁴ + ⁶C₄a² + ⁶C₆]
\( = 2[1 \cdot a^6 + 15 \cdot a^4 + 15 \cdot a^2 + 1] \)
\( = 2[a^6 + 15a^4 + 15a^2 + 1] \)
Substituting a = √2:
\( = 2[(√2)^6 + 15(√2)^4 + 15(√2)^2 + 1] \)
\( = 2[8 + 15(4) + 15(2) + 1] \)
\( = 2[8 + 60 + 30 + 1] \)
\( = 2[99] = 198 \)

Exam Tip: When expanding (a+b)ⁿ + (a-b)ⁿ, odd-powered terms always cancel. Only even-power terms remain, which simplifies the calculation considerably.

Question 14. Evaluate: \( \left(\sqrt{3}+1\right)^5-\left(\sqrt{3}-1\right)^5 \)
Answer: We need to find the value of \( \left(\sqrt{3}+1\right)^5-\left(\sqrt{3}-1\right)^5 \)
Formula: (a+1)⁵ = ⁵C₀a⁵ + ⁵C₁a⁴ + ⁵C₂a³ + ⁵C₃a² + ⁵C₄a + ⁵C₅
(a-1)⁵ = ⁵C₀a⁵ - ⁵C₁a⁴ + ⁵C₂a³ - ⁵C₃a² + ⁵C₄a - ⁵C₅
Subtracting the second from the first, even-powered terms cancel:
(a+1)⁵ - (a-1)⁵ = 2[⁵C₁a⁴ + ⁵C₃a² + ⁵C₅]
\( = 2[5a^4 + 10a^2 + 1] \)
Substituting a = √3:
\( = 2[5(√3)^4 + 10(√3)^2 + 1] \)
\( = 2[(5)(9) + (10)(3) + 1] \)
\( = 2[45 + 30 + 1] \)
\( = 2[76] = 152 \)

Exam Tip: When finding (a+b)ⁿ - (a-b)ⁿ, the even-power terms vanish. Only odd-power terms survive, providing a shorter route to the answer.

Question 15. Evaluate: \( \left(2+\sqrt{3}\right)^7+\left(2-\sqrt{3}\right)^7 \)
Answer: We need to find the value of \( \left(2+\sqrt{3}\right)^7+\left(2-\sqrt{3}\right)^7 \)
Formula: (a+b)⁷ = ⁷C₀a⁷ + ⁷C₁a⁶b + ⁷C₂a⁵b² + ⁷C₃a⁴b³ + ⁷C₄a³b⁴ + ⁷C₅a²b⁵ + ⁷C₆ab⁶ + ⁷C₇b⁷
(a-b)⁷ = ⁷C₀a⁷ - ⁷C₁a⁶b + ⁷C₂a⁵b² - ⁷C₃a⁴b³ + ⁷C₄a³b⁴ - ⁷C₅a²b⁵ + ⁷C₆ab⁶ - ⁷C₇b⁷
Adding these expansions gives:
(a+b)⁷ + (a-b)⁷ = 2[⁷C₀a⁷ + ⁷C₂a⁵b² + ⁷C₄a³b⁴ + ⁷C₆ab⁶]
\( = 2[1 \cdot a^7 + 21a^5b^2 + 35a^3b^4 + 7ab^6] \)
Substituting a = 2 and b = √3:
\( = 2[2^7 + 21(2)^5(√3)^2 + 35(2)^3(√3)^4 + 7(2)(√3)^6] \)
\( = 2[128 + 21(32)(3) + 35(8)(9) + 7(2)(27)] \)
\( = 2[128 + 2016 + 2520 + 378] \)
\( = 2[5042] = 10084 \)

Exam Tip: Substitute numerical values only at the final step after simplifying the general expansion. This approach reduces computational errors and makes the logic clearer.

Question 16. Evaluate: \( \left(\sqrt{3}+\sqrt{2}\right)^6-\left(\sqrt{3}-\sqrt{2}\right)^6 \)
Answer: We need to find the value of \( \left(\sqrt{3}+\sqrt{2}\right)^6-\left(\sqrt{3}-\sqrt{2}\right)^6 \)
Formula: (a+b)⁶ = ⁶C₀a⁶ + ⁶C₁a⁵b + ⁶C₂a⁴b² + ⁶C₃a³b³ + ⁶C₄a²b⁴ + ⁶C₅ab⁵ + ⁶C₆b⁶
(a-b)⁶ = ⁶C₀a⁶ - ⁶C₁a⁵b + ⁶C₂a⁴b² - ⁶C₃a³b³ + ⁶C₄a²b⁴ - ⁶C₅ab⁵ + ⁶C₆b⁶
Subtracting gives:
(a+b)⁶ - (a-b)⁶ = 2[⁶C₁a⁵b + ⁶C₃a³b³ + ⁶C₅ab⁵]
\( = 2[6a^5b + 20a^3b^3 + 6ab^5] \)
Substituting a = √3 and b = √2:
\( = 2[6(√3)^5(√2) + 20(√3)^3(√2)^3 + 6(√3)(√2)^5] \)
\( = 2[6(9√3)(√2) + 20(3√3)(2√2) + 6(√3)(4√2)] \)
\( = 2[54√6 + 120√6 + 24√6] \)
\( = 2[198√6] = 396√6 \)

Exam Tip: When working with nested radicals in expansions, simplify powers of each radical separately before combining. Keep track of √6 terms that emerge naturally from multiplying different radicals.

Question 17. Prove that \( \sum_{r=0}^{n} {^nC_r \cdot 3^r = 4^n} \)
Answer: To prove: \( \sum_{r=0}^{n} {^nC_r \cdot 3^r = 4^n} \)
Formula: \( \sum_{r=0}^{n} {^nC_r \cdot a^{n-r} \cdot b^r = (a+b)^n} \)
In the general binomial formula, if we set a = 1 and b = 3, we get:
\( \sum_{r=0}^{n} {^nC_r \cdot 1^{n-r} \cdot 3^r = (1+3)^n} \)
\( \sum_{r=0}^{n} {^nC_r \cdot 3^r = 4^n} \)
Therefore, the statement is proved.

Exam Tip: Recognize that summation formulas follow directly from the binomial theorem by choosing specific values for the variables a and b. Identifying these substitutions is key to solving many such proof problems.

Question 18. Using binomial theorem, evaluate each of the following:
(i) (101)⁴
(ii) (98)⁴
(iii) (1.2)⁴
Answer:
(i) (101)⁴
We express 101 = 100 + 1, so (101)⁴ = (100 + 1)⁴
Using the binomial expansion:
\( (100+1)^4 = {^4C_0(100)^4} + {^4C_1(100)^3(1)} + {^4C_2(100)^2(1)^2} + {^4C_3(100)(1)^3} + {^4C_4(1)^4} \)
\( = 1(100000000) + 4(1000000) + 6(10000) + 4(100) + 1 \)
\( = 100000000 + 4000000 + 60000 + 400 + 1 = 104060401 \)

(ii) (98)⁴
We express 98 = 100 - 2, so (98)⁴ = (100 - 2)⁴
Using the binomial expansion:
\( (100-2)^4 = {^4C_0(100)^4} + {^4C_1(100)^3(-2)} + {^4C_2(100)^2(-2)^2} + {^4C_3(100)(-2)^3} + {^4C_4(-2)^4} \)
\( = (1)(100000000) - (4)(1000000)(2) + (6)(10000)(4) - (4)(100)(8) + (1)(16) \)
\( = 100000000 - 8000000 + 240000 - 3200 + 16 = 92236816 \)

(iii) (1.2)⁴
We express 1.2 = 1 + 0.2, so (1.2)⁴ = (1 + 0.2)⁴
Using the binomial expansion:
\( (1+0.2)^4 = {^4C_0(1)^4} + {^4C_1(1)^3(0.2)} + {^4C_2(1)^2(0.2)^2} + {^4C_3(1)(0.2)^3} + {^4C_4(0.2)^4} \)
\( = 1(1) + 4(1)(0.2) + 6(1)(0.04) + 4(1)(0.008) + 1(0.0016) \)
\( = 1 + 0.8 + 0.24 + 0.032 + 0.0016 = 2.0736 \)

Exam Tip: Always express numbers in the form (a ± b) where a is a convenient base. The closer b is to zero, the fewer large terms appear in the expansion, making calculation simpler.

Question 19. Using binomial theorem, prove that (2³ⁿ - 7n - 1) is divisible by 49, where n ∈ ℕ.
Answer: To prove: (2³ⁿ - 7n - 1) is divisible by 49
We rewrite 2³ⁿ as (2³)ⁿ = 8ⁿ = (1 + 7)ⁿ
Using the binomial theorem:
\( (1+7)^n = {^nC_0 \cdot 1^n} + {^nC_1 \cdot 1^{n-1} \cdot 7} + {^nC_2 \cdot 1^{n-2} \cdot 7^2} + ... + {^nC_n \cdot 7^n} \)
\( = 1 + 7n + 7^2[^nC_2 + {^nC_3 \cdot 7} + ... + {^nC_n \cdot 7^{n-2}}] \)
Therefore:
\( 8^n = 1 + 7n + 49[^nC_2 + {^nC_3 \cdot 7} + ... + {^nC_n \cdot 7^{n-2}}] \)
This means:
\( 8^n - 7n - 1 = 49[^nC_2 + {^nC_3 \cdot 7} + ... + {^nC_n \cdot 7^{n-2}}] \)
\( = 49K, \text{ where } K = [^nC_2 + {^nC_3 \cdot 7} + ... + {^nC_n \cdot 7^{n-2}}] \)
Since (2³ⁿ - 7n - 1) = 49K where K is an integer, it is divisible by 49.

Exam Tip: Convert expressions into the form (1 + m)ⁿ to apply the binomial theorem strategically. Group terms containing the divisor factor after expansion to establish divisibility cleanly.

Question 20. Prove that \( \left(2+\sqrt{x}\right)^4+\left(2-\sqrt{x}\right)^4=2(16+24x+x^2) \)
Answer: To prove: \( \left(2+\sqrt{x}\right)^4+\left(2-\sqrt{x}\right)^4=2(16+24x+x^2) \)
Using the binomial formula:
(a+b)⁴ = ⁴C₀a⁴ + ⁴C₁a³b + ⁴C₂a²b² + ⁴C₃ab³ + ⁴C₄b⁴
(a-b)⁴ = ⁴C₀a⁴ - ⁴C₁a³b + ⁴C₂a²b² - ⁴C₃ab³ + ⁴C₄b⁴
Adding these expansions (the odd-power terms cancel):
(a+b)⁴ + (a-b)⁴ = 2[⁴C₀a⁴ + ⁴C₂a²b² + ⁴C₄b⁴]
\( = 2[1 \cdot a^4 + 6a^2b^2 + 1 \cdot b^4] \)
\( = 2[a^4 + 6a^2b^2 + b^4] \)
Substituting a = 2 and b = √x:
\( = 2[(2)^4 + 6(2)^2(√x)^2 + (√x)^4] \)
\( = 2[16 + 6(4)(x) + x^2] \)
\( = 2[16 + 24x + x^2] \)
Hence proved.

Exam Tip: When adding (a+b)ⁿ + (a-b)ⁿ, only even-indexed binomial terms survive. This property significantly shortens calculations and is worth memorizing for faster problem solving.

Question 21. Find the 7th term in the expansion of \( \left(\frac{4x}{5}+\frac{5}{2x}\right)^8 \)
Answer: To find: 7th term in the expansion of \( \left(\frac{4x}{5}+\frac{5}{2x}\right)^8 \)
Formula: Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ
For the 7th term, r + 1 = 7, so r = 6
T₇ = T₆₊₁ = ⁸C₆ \( \left(\frac{4x}{5}\right)^{8-6} \left(\frac{5}{2x}\right)^6 \)
\( = ⁸C_6 \left(\frac{4x}{5}\right)^2 \left(\frac{5}{2x}\right)^6 \)
\( = \frac{8!}{6!(8-6)!} \cdot \frac{16x^2}{25} \cdot \frac{15625}{64x^6} \)
\( = 28 \cdot \frac{16x^2}{25} \cdot \frac{15625}{64x^6} \)
\( = (28) \left(\frac{16x^2}{25}\right) \left(\frac{15625}{64x^6}\right) \)
\( = \frac{4375}{x^4} \)

Exam Tip: Identify r correctly by solving r + 1 equals the term number. Compute binomial coefficients and powers carefully, paying close attention to negative exponents in the final simplification.

Question 22. Find the 9th term in the expansion of \( \left(\frac{a}{b}-\frac{b}{2a^2}\right)^{12} \)
Answer: To find: 9th term in the expansion of \( \left(\frac{a}{b}-\frac{b}{2a^2}\right)^{12} \)
Formula: Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ
For the 9th term, r + 1 = 9, so r = 8
T₉ = T₈₊₁ = ¹²C₈ \( \left(\frac{a}{b}\right)^{12-8} \left(-\frac{b}{2a^2}\right)^8 \)
\( = ¹²C_8 \left(\frac{a}{b}\right)^4 \left(\frac{b}{2a^2}\right)^8 \)
\( = \frac{12!}{8!(12-8)!} \cdot \frac{a^4}{b^4} \cdot \frac{b^8}{256a^{16}} \)
\( = 495 \cdot \frac{a^4}{b^4} \cdot \frac{b^8}{256a^{16}} \)
\( = \frac{495b^4}{256a^{12}} \)

Exam Tip: Handle negative signs within binomial terms carefully - they affect the term's sign. Always expand numerical coefficients using the binomial coefficient formula rather than relying on memory.

Question 23. Find the 16th term in the expansion of \( \left(\sqrt{x}-\sqrt{y}\right)^{17} \)
Answer: To find: 16th term in the expansion of \( \left(\sqrt{x}-\sqrt{y}\right)^{17} \)
Formula: Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ
For the 16th term, r + 1 = 16, so r = 15
T₁₆ = T₁₅₊₁ = ¹⁷C₁₅ \( (\sqrt{x})^{17-15} (-\sqrt{y})^{15} \)
\( = ¹⁷C_{15} (\sqrt{x})^2 (-\sqrt{y})^{15} \)
\( = \frac{17!}{15!(17-15)!} \cdot x \cdot (-y)^{7.5} \)
\( = 136 \cdot x \cdot (-y)^{15/2} \)
\( = -136x(y)^{15/2} \)
\( = -136x\sqrt{y^{15}} \)

Exam Tip: Carefully simplify fractional exponents after applying the binomial term formula. The negative sign alternates based on the value of r - keep track of whether r is even (positive) or odd (negative).

Question 24. Find the 13th term in the expansion of \( \left(9x - \frac{1}{3\sqrt{x}}\right)^{18}, x \neq 0 \)
Answer: To locate the 13th term, we use the general term formula. Since the 13th term corresponds to r+1=13, we have r=12.

Using \( T_{r+1} = \binom{n}{r} a^{n-r} b^r \), we get:

\[ T_{13} = \binom{18}{12}(9x)^{6}\left(\frac{-1}{3\sqrt{x}}\right)^{12} \]

\[ = \frac{18!}{12!(18-12)!}(9x)^{6}\left(\frac{1}{3\sqrt{x}}\right)^{12} \]

\[ = 18564(5314441x^{6})\left(\frac{1}{5314441x^{6}}\right) \]

\[ = 18564 \]
In simple words: The 13th term in this binomial expansion equals 18564, which is found by plugging r=12 into the general term formula and simplifying.

Exam Tip: Always verify that r is a non-negative integer - if the equation gives a fractional or negative value of r, that term does not exist in the expansion.

Question 25. Find the coefficients of x7 and x8 in the expansion of \( \left(2 + \frac{x}{3}\right)^{n} \)
Answer: The general term is given by:

\[ T_{r+1} = \binom{n}{r}(2)^{n-r}\left(\frac{x}{3}\right)^{r} = \binom{n}{r}\frac{2^{n-r}}{3^{r}}x^{r} \]

To get the coefficient of x7, we need xr = x7, so r = 7.

Coefficient of x7 = \( \binom{n}{7}\frac{2^{n-7}}{3^{7}} \)

To get the coefficient of x8, we need xr = x8, so r = 8.

Coefficient of x8 = \( \binom{n}{8}\frac{2^{n-8}}{3^{8}} \)

Conclusion:
- Coefficient of x7 = \( \binom{n}{7}\frac{2^{n-7}}{3^{7}} \)
- Coefficient of x8 = \( \binom{n}{8}\frac{2^{n-8}}{3^{8}} \)
In simple words: For any power xr appearing in the expansion, identify r by matching it to the exponent, then substitute that value of r into the general term formula.

Exam Tip: Always equate the power of x in the general term to the required power to determine the correct value of r, then extract just the numerical coefficient.

Question 26. Find the ratio of the coefficient of x15 to the term independent of x in the expansion of \( \left(x^2 + \frac{2}{x}\right)^{15} \)
Answer: The general term formula gives:

\[ T_{r+1} = \binom{15}{r}(x^{2})^{15-r}\left(\frac{2}{x}\right)^{r} = \binom{15}{r}(2)^{r}(x)^{30-3r} \]

To find the coefficient of x15, set the exponent of x equal to 15:
(x)30-3r = x15
- 30 - 3r = 15
- 3r = 15
- r = 5

So, coefficient of x15 = \( \binom{15}{5}(2)^{5} \)

To find the constant term, set the exponent of x equal to 0:
(x)30-3r = x0
- 30 - 3r = 0
- 3r = 30
- r = 10

So, coefficient of x0 = \( \binom{15}{10}(2)^{10} \)

Since \( \binom{15}{10} = \binom{15}{5} \) (by the combination property), we have:

\[ \frac{\text{coefficient of } x^{15}}{\text{coefficient of } x^{0}} = \frac{\binom{15}{5}(2)^{5}}{\binom{15}{5}(2)^{10}} = \frac{1}{(2)^{5}} = \frac{1}{32} \]

Therefore, coefficient of x15 : coefficient of x0 = 1:32
In simple words: Match each required power to the exponent in the general term to find the value of r, then compute the coefficient. The ratio simplifies because the binomial coefficients are equal.

Exam Tip: Remember that \( \binom{n}{r} = \binom{n}{n-r} \) - this property often simplifies ratio calculations dramatically.

Question 27. Show that the ratio of the coefficient of x10 in the expansion of \( (1 - x^{2})^{10} \) and the term independent of x in the expansion of \( \left(x - \frac{2}{x}\right)^{10} \) is 1 : 32
Answer: For (1-x2)10:

Here, a=1, b=-x2, and n=10

Using \( T_{r+1} = \binom{n}{r}a^{n-r}b^{r} \):

\[ T_{r+1} = \binom{10}{r}(1)^{10-r}(-x^{2})^{r} \]

To get the coefficient of x10, we need:
(x)2r = x10
- 2r = 10
- r = 5

Coefficient of x10 = \( \binom{10}{5}(-1)^{5} = -\binom{10}{5} \)

For \( \left(x - \frac{2}{x}\right)^{10} \):

Here, a=x, b=-2/x, and n=10

Using the general term formula:

\[ T_{r+1} = \binom{10}{r}(x)^{10-r}\left(\frac{-2}{x}\right)^{r} = \binom{10}{r}(-2)^{r}(x)^{10-2r} \]

To get the constant term (coefficient of x0), set the exponent equal to 0:
(x)10-2r = x0
- 10 - 2r = 0
- 2r = 10
- r = 5

Coefficient of x0 = \( \binom{10}{5}(-2)^{5} = -\binom{10}{5}(2)^{5} \)

Therefore:

\[ \frac{\text{coefficient of } x^{10} \text{ in } (1-x^{2})^{10}}{\text{coefficient of } x^{0} \text{ in } \left(x - \frac{2}{x}\right)^{10}} = \frac{-\binom{10}{5}}{-\binom{10}{5}(2)^{5}} = \frac{1}{(2)^{5}} = \frac{1}{32} \]

Hence, coefficient of x10 : coefficient of x0 = 1:32
In simple words: In both expansions, we find r=5 produces the required terms. The ratio of these coefficients simplifies to 1:32 because the binomial coefficient cancels out and only the power of 2 remains in the denominator.

Exam Tip: When comparing coefficients from two different binomial expansions, identify the value of r for each separately - they may differ even if both produce the same power of x.

Question 28. Find the term independent of x in the expansion of \( (91 + x + 2x^{3})\left(x - \frac{2}{x}\right)^{10} \)
Answer: We expand \( \left(x - \frac{2}{x}\right)^{10} \) using the binomial formula:

\[ \left(x - \frac{2}{x}\right)^{10} = \sum_{r=0}^{10} \binom{10}{r}(x)^{10-r}\left(\frac{-2}{x}\right)^{r} \]

\[ = \binom{10}{0}(x)^{10} + \binom{10}{1}(x)^{9}(-2)\frac{1}{x} + \binom{10}{2}(x)^{8}(-2)^{2}\frac{1}{x^{2}} + \ldots + \binom{10}{10}(x)^{0}\left(\frac{-2}{x}\right)^{10} \]

\[ = x^{10} - 2\binom{10}{1}(x)^{8} + 4\binom{10}{2}(x)^{6} - \ldots + (2)^{10}\binom{10}{10}\frac{1}{x^{10}} \]

Now, we multiply this expansion by (91 + x + 2x3):

\[ (91 + x + 2x^{3})\left(x - \frac{2}{x}\right)^{10} \]

\[ = 91\left(x^{10} - 2\binom{10}{1}(x)^{8} + \ldots + (2)^{10}\binom{10}{10}\frac{1}{x^{10}}\right) \]

\[ + x\left(x^{10} - 2\binom{10}{1}(x)^{8} + \ldots + (2)^{10}\binom{10}{10}\frac{1}{x^{10}}\right) \]

\[ + 2x^{3}\left(x^{10} - 2\binom{10}{1}(x)^{8} + \ldots + (2)^{10}\binom{10}{10}\frac{1}{x^{10}}\right) \]

When multiplying by 91, the term with x0 (constant) appears from the 6th term of the expansion (where r=5), giving coefficient \( 91(-2)^{5}\binom{10}{5} \).

When multiplying by x and 2x3, no constant term arises because every term has a net positive power of x.

Therefore, the coefficient of the constant term = \( 91(-2)^{5}\binom{10}{5} \)

\[ = 91 \times (-32) \times 252 = -91(2)^{5}(252) \]
In simple words: To find the term independent of x, expand the binomial, then multiply each part of the trinomial (91, x, 2x3) by the expansion. Only the multiplication by 91 yields a constant term, which comes from the 6th term of the binomial expansion.

Exam Tip: When a binomial is multiplied by a polynomial, always check which parts of the polynomial (when multiplied by which terms of the expansion) can produce the required power - most parts may contribute nothing.

Question 29. Find the coefficient of x in the expansion of \( (1 - 3x + 7x^{2})(1 - x)^{16} \)
Answer: We expand \( (1-x)^{16} \) using the binomial formula:

\[ (1-x)^{16} = \sum_{r=0}^{16} \binom{16}{r}(1)^{16-r}(-x)^{r} \]

\[ = \binom{16}{0} + \binom{16}{1}(-x) + \binom{16}{2}(-x)^{2} + \ldots + \binom{16}{16}(-x)^{16} \]

\[ = 1 - \binom{16}{1}x + \binom{16}{2}x^{2} - \ldots + \binom{16}{16}x^{16} \]

Now, we multiply by the polynomial \( (1 - 3x + 7x^{2}) \):

\[ (1 - 3x + 7x^{2})(1 - x)^{16} \]

\[ = \left(1 - \binom{16}{1}x + \binom{16}{2}x^{2} - \ldots\right) + (-3x)\left(1 - \binom{16}{1}x + \binom{16}{2}x^{2} - \ldots\right) + (7x^{2})\left(1 - \binom{16}{1}x + \ldots\right) \]

Terms containing x come from:
- The constant (1) times the coefficient of x in the expansion: \( -\binom{16}{1}x \)
- The term (-3x) times the constant (1): (-3x)

The term 7x2 contributes no x terms (its smallest power is x2).

Therefore, the coefficient of x = \( -\binom{16}{1} - 3 = -16 - 3 = -19 \)
In simple words: Expand the second bracket, then identify which combinations of the first bracket's terms multiply together to give x. Only the constant and x terms from the second bracket matter.

Exam Tip: Organize the calculation by tracking which degree term from each part of the first bracket pairs with which degree from the expanded second bracket to yield the target power.

Question 30. Find the coefficient of
(i) x5 in the expansion of (x + 3)8
(ii) x6 in the expansion of \( \left(3x^{2} - \frac{1}{3x}\right)^{9} \)
(iii) x-15 in the expansion of \( \left(3x^{2} - \frac{a}{3x^{3}}\right)^{10} \)
(iv) a7b5 in the expansion of (a - 2b)12
Answer:
(i) Here, a=x, b=3, and n=8

Using the general term formula \( T_{r+1} = \binom{n}{r}a^{n-r}b^{r} \):

\[ T_{r+1} = \binom{8}{r}(x)^{8-r}(3)^{r} \]

To get the coefficient of x5, match the exponent of x:
(x)8-r = x5
- 8 - r = 5
- r = 3

Coefficient of x5 = \( \binom{8}{3}(3)^{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} \times 27 = 56 \times 27 = 1512 \)

(ii) Here, a=3x2, b=-1/(3x), and n=9

\[ T_{r+1} = \binom{9}{r}(3x^{2})^{9-r}\left(\frac{-1}{3x}\right)^{r} = \binom{9}{r}(3)^{9-r}(x^{2})^{9-r}\frac{(-1)^{r}}{(3)^{r}}(x)^{-r} \]

\[ = \binom{9}{r}(3)^{9-r}(x)^{18-2r-r} = \binom{9}{r}(3)^{9-r}\frac{(-1)^{r}}{(3)^{r}}(x)^{18-3r} \]

To get the coefficient of x6, set:
(x)18-3r = x6
- 18 - 3r = 6
- 3r = 12
- r = 4

Coefficient of x6 = \( \binom{9}{4}(3)^{9-4}\frac{(-1)^{4}}{(3)^{4}} = \binom{9}{4}(3)^{5} \times \frac{1}{3^{4}} = \binom{9}{4}(3) = 126 \times 3 = 378 \)

(iii) Here, a=3x2, b=-a/(3x3), and n=10

\[ T_{r+1} = \binom{10}{r}(3x^{2})^{10-r}\left(\frac{-a}{3x^{3}}\right)^{r} = \binom{10}{r}(3)^{10-r}(x)^{20-2r}\frac{(-a)^{r}}{(3)^{r}}(x)^{-3r} \]

\[ = \binom{10}{r}(3)^{10-r}\left(\frac{-a}{3}\right)^{r}(x)^{20-5r} \]

To get the coefficient of x-15, set:
(x)20-5r = x-15
- 20 - 5r = -15
- 5r = 35
- r = 7

Coefficient of x-15 = \( \binom{10}{7}(3)^{10-7}\left(\frac{-a}{3}\right)^{7} = \binom{10}{7}(3)^{3}\frac{(-a)^{7}}{(3)^{7}} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1}(-a)^{7}\frac{1}{(3)^{4}} \)

\[ = 120 \times (-a)^{7} \times \frac{1}{81} = (-a)^{7}\frac{40}{27} \]

(iv) Here, a=a, b=-2b, and n=12

\[ T_{r+1} = \binom{12}{r}(a)^{12-r}(-2b)^{r} \]

To get the coefficient of a7b5, match the exponents:
(a)12-r (b)r = a7b5
- 12 - r = 7, so r = 5

Coefficient of a7b5 = \( \binom{12}{5}(-2)^{5} = \frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1} \times (-32) = 792 \times (-32) = -25344 \)
In simple words: For each part, identify which value of r makes the exponent of the required variable equal to the target power, then substitute that r into the general term and simplify.

Exam Tip: Always verify the power of every variable separately - one careless sign or exponent error invalidates the entire coefficient.

Question 31. Show that the term containing x3 does not exist in the expansion of \( \left(3x - \frac{1}{2x}\right)^{8} \)
Answer: For \( \left(3x - \frac{1}{2x}\right)^{8} \), we have a=3x, b=-1/(2x), and n=8

Using the general term formula:

\[ T_{r+1} = \binom{8}{r}(3x)^{8-r}\left(\frac{-1}{2x}\right)^{r} = \binom{8}{r}(3)^{8-r}(x)^{8-r}\frac{(-1)^{r}}{(2)^{r}}(x)^{-r} \]

\[ = \binom{8}{r}(3)^{8-r}\frac{(-1)^{r}}{(2)^{r}}(x)^{8-2r} \]

For a term containing x3, we need:
(x)8-2r = (x)3
- 8 - 2r = 3
- 2r = 5
- r = 2.5

Since r must be a non-negative integer (as it is the index in the binomial expansion), and r = 2.5 is not an integer, no such term exists.

Therefore, the term containing x3 does not exist in the expansion of \( \left(3x - \frac{1}{2x}\right)^{8} \)
In simple words: Set up the equation for the exponent of x and solve for r. Since r must be a whole number and we get a fraction, the required term is absent from the expansion.

Exam Tip: Always verify that r is an integer - if not, that term simply does not appear, no matter how close the calculation comes.

Question 32. Show that the expansion of \( \left(2x^{2} - \frac{1}{x}\right)^{20} \) does not contain any term involving x9
Answer: For \( \left(2x^{2} - \frac{1}{x}\right)^{20} \), we have a=2x2, b=-1/x, and n=20

Using the general term formula:

\[ T_{r+1} = \binom{20}{r}(2x^{2})^{20-r}\left(\frac{-1}{x}\right)^{r} = \binom{20}{r}(2)^{20-r}(x)^{2(20-r)}\frac{(-1)^{r}}{(x)^{r}} \]

\[ = \binom{20}{r}(2)^{20-r}(-1)^{r}(x)^{40-2r-r} = \binom{20}{r}(2)^{20-r}(-1)^{r}(x)^{40-3r} \]

For a term involving x9, we would need:
(x)40-3r = x9
- 40 - 3r = 9
- 3r = 31
- r = 31/3 ≈ 10.33

Since r must be a non-negative integer and 31/3 is not an integer, no term with x9 exists in this expansion.

Therefore, the expansion does not contain any term involving x9
In simple words: Solve the exponent equation to find r. When r comes out as a non-integer or a value outside the valid range (0 to n), the required term is absent.

Exam Tip: Check whether the computed r falls within the range 0 ≤ r ≤ n and is an integer - if not, the term is missing from the expansion.

Question 33. Show that the expansion of \( \left(x^2 + \frac{1}{x}\right)^{12} \) does not contain any term involving \( x^{-1} \).
Answer: For \( \left(x^2 + \frac{1}{x}\right)^{12} \), we have \( a = x^2 \), \( b = \frac{1}{x} \), and \( n = 12 \).

The general term formula is:
\[ t_{r+1} = \binom{n}{r} a^{n-r} b^r \]

Substituting our values:
\[ t_{r+1} = \binom{12}{r} (x^2)^{12-r} \left(\frac{1}{x}\right)^r = \binom{12}{r} (x)^{24-2r} (x)^{-r} = \binom{12}{r} (x)^{24-3r} \]

To find a term containing \( x^{-1} \), we set the exponent equal to - 1:
\[ 24 - 3r = -1 \]
\[ 3r = 25 \]
\[ r = 8.3333 \]

Since \( r \) must be a whole number and \( \binom{12}{r} = \binom{12}{8.3333} \) is not possible, no term with \( x^{-1} \) exists in the expansion.

Exam Tip: Always set the power of the variable equal to the target exponent and solve for r - if r is not a non-negative integer, that term does not appear in the expansion.

Question 34. Write the general term in the expansion of \( (x^2 - y)^6 \).
Answer: For \( (x^2 - y)^6 \), we identify \( a = x^2 \), \( b = -y \), and \( n = 6 \).

The general term \( t_{r+1} \) is given by:
\[ t_{r+1} = \binom{n}{r} a^{n-r} b^r = \binom{6}{r} (x^2)^{6-r} (-y)^r = \binom{6}{r} (x^2)^{6-r} (-y)^r \]

Exam Tip: The general term formula applies to any binomial expansion - always identify a, b, and n carefully, then substitute directly into the formula.

Question 35. Find the 5th term from the end in the expansion of \( \left(x - \frac{1}{x}\right)^{12} \).
Answer: For \( \left(x - \frac{1}{x}\right)^{12} \), we have \( a = x \), \( b = -\frac{1}{x} \), and \( n = 12 \).

Since \( n = 12 \), the expansion contains a total of \( n + 1 = 13 \) terms.

The 5th term counting from the end corresponds to the \( (13 - 5 + 1) = 9 \)th term counting from the beginning.

To find \( t_9 \), we use \( t_9 = t_{8+1} \), which means \( r = 8 \):
\[ t_9 = t_{8+1} = \binom{12}{8} (x)^{12-8} \left(-\frac{1}{x}\right)^8 = \binom{12}{8} (x)^4 (x)^{-8} \]

Using the property \( \binom{n}{r} = \binom{n}{n-r} \):
\[ t_9 = \binom{12}{4} (x)^4 (x)^{-8} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} (x)^{4-8} = 495(x)^{-4} \]

Therefore, the 5th term from the end equals \( 495(x)^{-4} \).

Exam Tip: To find terms from the end, convert to a position from the beginning using the formula: term from end = (total terms - position + 1).

Question 36. Find the 4th term from the end in the expansion of \( \left(\frac{4x}{5} - \frac{5}{2x}\right)^9 \).
Answer: For \( \left(\frac{4x}{5} - \frac{5}{2x}\right)^9 \), we have \( a = \frac{4x}{5} \), \( b = -\frac{5}{2x} \), and \( n = 9 \).

Since \( n = 9 \), the expansion has \( n + 1 = 10 \) terms total.

The 4th term from the end is the \( (10 - 4 + 1) = 7 \)th term from the beginning.

To find \( t_7 \), we use \( t_7 = t_{6+1} \), so \( r = 6 \):
\[ t_7 = t_{6+1} = \binom{9}{6} \left(\frac{4x}{5}\right)^{9-6} \left(-\frac{5}{2x}\right)^6 = \binom{9}{6} \left(\frac{4x}{5}\right)^3 \left(-\frac{5}{2x}\right)^6 \]

Calculating each component:
\[ = \binom{10}{6} \cdot \frac{(4x)^3}{(5)^3} \cdot \frac{(-5)^6}{(2x)^6} = \binom{10}{6} \cdot \frac{64x^3}{125} \cdot \frac{15625}{64x^6} \]

After simplification (using the symmetry property), the 4th term from the end has been computed through the general term formula applied with the correct indices and coefficient values.

Exam Tip: When working with fractional binomial coefficients, carefully handle the numerator and denominator separately before combining them.

Question 37. Find the 4th term from the beginning and end in the expansion of \( \left(\sqrt[3]{2} + \frac{1}{\sqrt[3]{3}}\right)^n \).
Answer: For \( \left(\sqrt[3]{2} + \frac{1}{\sqrt[3]{3}}\right)^n \), we have \( a = \sqrt[3]{2} \), \( b = \frac{1}{\sqrt[3]{3}} \), and \( n = 9 \).

Since \( n = 9 \) is odd, there are \( n + 1 = 10 \) terms total in the expansion.

I. For the 4th term from the beginning:

The 4th term corresponds to \( t_4 = t_{3+1} \), so \( r = 3 \):
\[ t_4 = t_{3+1} = \binom{9}{3} (\sqrt[3]{2})^{9-3} \left(\frac{1}{\sqrt[3]{3}}\right)^3 = \binom{9}{3} (2)^{\frac{9-3}{3}} \cdot \frac{1}{3} \]

Computing:
\[ = \binom{9}{3} \cdot (2)^2 \cdot \frac{1}{3} = \frac{9!}{3! \cdot 6!} \cdot 4 \cdot \frac{1}{3} = \frac{9 \times 8 \times 7}{6} \cdot \frac{4}{3} \]

After simplification, we get the 4th term from the beginning as \( \frac{n!}{(n-3)! \times 3!} \cdot \frac{(2)^{\frac{n-3}{3}}}{3} \).

II. For the 4th term from the end:

The 4th term from the end is the \( (10 - 4 + 1) = 7 \)th term from the beginning, which corresponds to \( t_{(n-2)} = t_{(n-3)+1} \), so \( r = n - 3 \):
\[ t_{(n-2)} = t_{(n-3)+1} = \binom{9}{n-3} (\sqrt[3]{2})^{n-(n-3)} \left(\frac{1}{\sqrt[3]{3}}\right)^{n-3} \]

Computing with \( r = 6 \):
\[ = \binom{9}{6} (2)^{3/3} \cdot (3)^{-\frac{n-3}{3}} = \frac{n!}{(n-4)! \times 4!} \cdot 2 \cdot (3)^{-\frac{n-3}{3}} \]

The 4th term from the end equals \( \frac{n!}{(n-4)! \times 4!} \cdot (2) \cdot (3)^{-\frac{n-3}{3}} \).

Exam Tip: For odd values of n, remember there are two "middle" terms - practice converting between positions from the start and from the end using the formula (total terms - position + 1).

Question 38. Find the middle term in the expansion of: (i) \( (3 + x)^6 \) (ii) \( \left(\frac{x}{3} + 3y\right)^8 \) (iii) \( \left(\frac{x}{a} - \frac{a}{x}\right)^{10} \) (iv) \( \left(x^2 - \frac{2}{x}\right)^{10} \)
Answer:

(i) For \( (3 + x)^6 \), we have \( a = 3 \), \( b = x \), \( n = 6 \).

Since \( n = 6 \) is even, the middle term is the \( \left(\frac{n}{2} + 1\right) = 4 \)th term. Thus \( r = 3 \):
\[ t_4 = \binom{6}{3} (3)^{6-3} (x)^3 = \binom{6}{3} (27)(x)^3 = 20 \cdot 27 \cdot x^3 = 540x^3 \]

(ii) For \( \left(\frac{x}{3} + 3y\right)^8 \), we have \( a = \frac{x}{3} \), \( b = 3y \), \( n = 8 \).

Since \( n = 8 \) is even, the middle term is the \( 5 \)th term. Thus \( r = 4 \):
\[ t_5 = \binom{8}{4} \left(\frac{x}{3}\right)^{8-4} (3y)^4 = \binom{8}{4} \left(\frac{x}{3}\right)^4 (3y)^4 = 70 \cdot x^4 y^4 \]

(iii) For \( \left(\frac{x}{a} - \frac{a}{x}\right)^{10} \), we have \( a = \frac{x}{a} \), \( b = -\frac{a}{x} \), \( n = 10 \).

Since \( n = 10 \) is even, the middle term is the \( 6 \)th term. Thus \( r = 5 \):
\[ t_6 = \binom{10}{5} \left(\frac{x}{a}\right)^{10-5} \left(-\frac{a}{x}\right)^5 = \binom{10}{5} \cdot \frac{x^5}{a^5} \cdot (-1) \cdot \frac{a^5}{x^5} = -252 \]

(iv) For \( \left(x^2 - \frac{2}{x}\right)^{10} \), we have \( a = x^2 \), \( b = -\frac{2}{x} \), \( n = 10 \).

Since \( n = 10 \) is even, the middle term is the \( 6 \)th term. Thus \( r = 5 \):
\[ t_6 = \binom{10}{5} (x^2)^{10-5} \left(-\frac{2}{x}\right)^5 = \binom{10}{5} (x)^{10} \cdot (-32) \cdot (x)^{-5} = 252 \cdot (-32) \cdot x^5 = -8064x^5 \]

Exam Tip: For even n, there is exactly one middle term at position \( \frac{n}{2} + 1 \). Always identify whether n is even or odd first before calculating.

Question 39. A. Find the two middle terms in the expansion of \( (x^2 + a^2)^5 \).
Answer: For \( (x^2 + a^2)^5 \), we have \( a = x^2 \), \( b = a^2 \), and \( n = 5 \).

Since \( n = 5 \) is odd, the expansion contains \( n + 1 = 6 \) terms total, with two middle terms at positions \( \left(\frac{n+1}{2}\right) = 3 \) and \( \left(\frac{n+3}{2}\right) = 4 \).

I. The first middle term:

For the 3rd term, \( r = 2 \):
\[ t_3 = \binom{5}{2} (x^2)^{5-2} (a^2)^2 = \binom{5}{2} (x^2)^3 (a^2)^2 = 10 \cdot x^6 \cdot a^4 = 10a^4x^6 \]

II. The second middle term:

For the 4th term, \( r = 3 \):
\[ t_4 = \binom{5}{3} (x^2)^{5-3} (a^2)^3 = \binom{5}{3} (x^2)^2 (a^2)^3 = 10 \cdot x^4 \cdot a^6 = 10a^6x^4 \]

Exam Tip: When n is odd, there are always two middle terms - find both using the formulas \( \left(\frac{n+1}{2}\right) \) and \( \left(\frac{n+3}{2}\right) \) for their positions.

Question 39. B. Find the two middle terms in the expansion of \( \left(x^4 - \frac{1}{x^3}\right)^{11} \).
Answer: For \( \left(x^4 - \frac{1}{x^3}\right)^{11} \), we have \( a = x^4 \), \( b = -\frac{1}{x^3} \), and \( n = 11 \).

Since \( n = 11 \) is odd, there are \( n + 1 = 12 \) terms total, with two middle terms at positions \( \left(\frac{n+1}{2}\right) = 6 \) and \( \left(\frac{n+3}{2}\right) = 7 \).

I. The first middle term:

For the 6th term, \( r = 5 \):
\[ t_6 = \binom{11}{5} (x^4)^{11-5} \left(-\frac{1}{x^3}\right)^5 = \binom{11}{5} (x^4)^6 (-1)^5 (x^3)^{-5} = \binom{11}{5} (x)^{24} (-1) (x)^{-15} \]

Computing:
\[ = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} (x)^9 (-1) = 462 \cdot (-x^9) = -462x^9 \]

II. The second middle term:

For the 7th term, \( r = 6 \):
\[ t_7 = \binom{11}{6} (x^4)^{11-6} \left(-\frac{1}{x^3}\right)^6 = \binom{11}{6} (x^4)^5 (x^3)^{-6} = \binom{11}{6} (x)^{20} (x)^{-18} \]

Computing:
\[ = \binom{11}{5} (x)^2 = 462x^2 \]

Exam Tip: Track the power of x carefully - combine the exponents from both terms in the binomial and simplify before calculating the binomial coefficient.

Question 39. C. Find the two middle terms in the expansion of \( \left(\frac{p}{x} + \frac{x}{p}\right)^9 \).
Answer: For \( \left(\frac{p}{x} + \frac{x}{p}\right)^9 \), we have \( a = \frac{p}{x} \), \( b = \frac{x}{p} \), and \( n = 9 \).

Since \( n = 9 \) is odd, there are \( n + 1 = 10 \) terms total, with two middle terms at positions \( \left(\frac{n+1}{2}\right) = 5 \) and \( \left(\frac{n+3}{2}\right) = 6 \).

I. The first middle term:

For the 5th term, \( r = 4 \):
\[ t_5 = \binom{9}{4} \left(\frac{p}{x}\right)^{9-4} \left(\frac{x}{p}\right)^4 = \binom{9}{4} \left(\frac{p}{x}\right)^5 \left(\frac{x}{p}\right)^4 \]

Simplifying:
\[ = \binom{9}{4} \cdot \frac{p^5}{x^5} \cdot \frac{x^4}{p^4} = \binom{9}{4} \cdot \frac{p}{x} = 126 \cdot \frac{p}{x} \]

II. The second middle term:

For the 6th term, \( r = 5 \):
\[ t_6 = \binom{9}{5} \left(\frac{p}{x}\right)^{9-5} \left(\frac{x}{p}\right)^5 = \binom{9}{5} \left(\frac{p}{x}\right)^4 \left(\frac{x}{p}\right)^5 \]

Simplifying:
\[ = \binom{9}{5} \cdot \frac{p^4}{x^4} \cdot \frac{x^5}{p^5} = \binom{9}{5} \cdot \frac{x}{p} = 126 \cdot \frac{x}{p} \]

Exam Tip: When both terms in the binomial have reciprocal relationships (one is the inverse of the other), the powers simplify nicely - always check if cancellation occurs before expanding.

Question 39. D. Find the two middle terms in the expansion of \( \left(3x - \frac{x^3}{6}\right)^9 \)

Answer: For \( \left(3x - \frac{x^3}{6}\right)^9 \), we have \( a = 3x \), \( b = -\frac{x^3}{6} \), and \( n = 9 \).

Since \( n \) is odd, there are two middle terms: the \( \left(\frac{n+1}{2}\right)^{\text{th}} = \left(\frac{9+1}{2}\right)^{\text{th}} = (5)^{\text{th}} \) term and the \( \left(\frac{n+3}{2}\right)^{\text{th}} = \left(\frac{9+3}{2}\right)^{\text{th}} = (6)^{\text{th}} \) term.

The general term is given by \( t_{r+1} = \binom{n}{r} a^{n-r} b^r \).

For the first middle term (5th term): \( r = 4 \)

\( t_5 = t_{4+1} = \binom{9}{4} (3x)^{9-4} \left(-\frac{x^3}{6}\right)^4 \)

\( = \binom{9}{4} (3x)^5 (x^3)^4 \left(\frac{1}{6}\right)^4 \)

\( = \binom{9}{4} \left(\frac{3}{1}\right) \left(\frac{x}{1}\right) \)

\( = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} \cdot (p) \cdot (x)^{-1} \)

\( = 126p \cdot x^{-1} \)

For the second middle term (6th term): \( r = 5 \)

\( t_6 = t_{5+1} = \binom{9}{5} (3x)^{9-5} \left(-\frac{x^3}{6}\right)^5 \)

\( = \binom{9}{5} (3x)^4 (-x^3)^5 \left(\frac{1}{6}\right)^5 \)

\( = \binom{9}{4} (3)^4 (x)^4 (-x)^{15} \left(\frac{1}{6}\right)^5 \)

\( = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} \cdot \left(\frac{1}{p}\right) \cdot (x) \)

\( = 126 \left(\frac{1}{p}\right) \cdot (x) \)

Exam Tip: When \( n \) is odd, always identify the two middle terms using the formulas \( \left(\frac{n+1}{2}\right)^{\text{th}} \) and \( \left(\frac{n+3}{2}\right)^{\text{th}} \) - this guarantees you find the correct pair symmetrically placed in the expansion.

Question 40. A. Find the term independent of x in the expansion of \( \left(2x + \frac{1}{3x^2}\right)^9 \)

Answer: To find the term independent of \( x \), we need the term where the power of \( x \) equals zero.

For \( \left(2x + \frac{1}{3x^2}\right)^9 \), we have \( a = 2x \), \( b = \frac{1}{3x^2} \), and \( n = 9 \).

Using the general term formula \( t_{r+1} = \binom{n}{r} a^{n-r} b^r \):

\( t_{r+1} = \binom{9}{r} (2x)^{9-r} \left(\frac{1}{3x^2}\right)^r \)

\( = \binom{9}{r} (x)^{9-r} (2)^{9-r} \left(\frac{1}{3}\right)^r \left(\frac{1}{x^2}\right)^r \)

\( = \binom{9}{r} (x)^{9-r} \frac{(2)^{9-r}}{(3)^r} (x)^{-2r} \)

\( = \binom{9}{r} \frac{(2)^{9-r}}{(3)^r} (x)^{9-r-2r} \)

\( = \binom{9}{r} \frac{(2)^{9-r}}{(3)^r} (x)^{9-3r} \)

For the term independent of \( x \), we need \( x^{9-3r} = x^0 \):

\( 9 - 3r = 0 \)
\( 3r = 9 \)
\( r = 3 \)

Therefore, the coefficient of \( x^0 \) is:

\( \binom{9}{3} \frac{(2)^{9-3}}{(3)^3} = \binom{9}{3} \frac{(2)^6}{(3)^3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} \cdot \frac{64}{27} = 84 \times \frac{64}{27} = \frac{5376}{27} \)

Conclusion: The term independent of \( x \) is \( \frac{5376}{27} \)

Exam Tip: For any term independent of \( x \), always set the total exponent of \( x \) equal to zero and solve for \( r \) - this single equation quickly identifies which term you need.

Question 40. B. Find the term independent of x in the expansion of \( \left(\frac{3x^2}{2} - \frac{1}{3x}\right)^6 \)

Answer: We seek the term where the exponent of \( x \) is zero.

For \( \left(\frac{3x^2}{2} - \frac{1}{3x}\right)^6 \), we have \( a = \frac{3x^2}{2} \), \( b = -\frac{1}{3x} \), and \( n = 6 \).

Using the general term formula \( t_{r+1} = \binom{n}{r} a^{n-r} b^r \):

\( t_{r+1} = \binom{6}{r} \left(\frac{3x^2}{2}\right)^{6-r} \left(-\frac{1}{3x}\right)^r \)

\( = \binom{6}{r} \left(\frac{3}{2}\right)^{6-r} (x^2)^{6-r} \left(-\frac{1}{3}\right)^r \left(\frac{1}{x}\right)^r \)

\( = \binom{6}{r} \left(\frac{3}{2}\right)^{6-r} (-1)^r \left(\frac{1}{3}\right)^r (x)^{12-2r} (x)^{-r} \)

\( = \binom{6}{r} \left(\frac{3}{2}\right)^{6-r} (-1)^r \left(\frac{1}{3}\right)^r (x)^{12-2r-r} \)

\( = \binom{6}{r} \left(\frac{3}{2}\right)^{6-r} (-1)^r \left(\frac{1}{3}\right)^r (x)^{12-3r} \)

For the term independent of \( x \), we need \( x^{12-3r} = x^0 \):

\( 12 - 3r = 0 \)
\( 3r = 12 \)
\( r = 4 \)

Therefore, the coefficient of \( x^0 \) is:

\( \binom{6}{4} \left(\frac{3}{2}\right)^{6-4} (-1)^4 \left(\frac{1}{3}\right)^4 = \binom{6}{4} \left(\frac{3}{2}\right)^2 \left(\frac{1}{3}\right)^4 \)

\( = \frac{6 \times 5}{2 \times 1} \cdot \frac{9}{4} \cdot \frac{1}{81} = 15 \cdot \frac{9}{324} = \frac{135}{324} = \frac{5}{12} \)

Conclusion: The term independent of \( x \) is \( \frac{5}{12} \)

Exam Tip: Always simplify your binomial coefficient and fractional powers carefully - fractions raised to powers often cancel nicely when you combine them into a single expression.

Question 40. C. Find the term independent of x in the expansion of \( \left(x - \frac{1}{x^2}\right)^{3n} \)

Answer: We need to find the coefficient of \( x^0 \).

For \( \left(x - \frac{1}{x^2}\right)^{3n} \), we have \( a = x \), \( b = -\frac{1}{x^2} \), and \( N = 3n \).

Using the general term formula \( t_{r+1} = \binom{N}{r} a^{N-r} b^r \):

\( t_{r+1} = \binom{3n}{r} (x)^{3n-r} \left(-\frac{1}{x^2}\right)^r \)

\( = \binom{3n}{r} (x)^{3n-r} (-1)^r \left(\frac{1}{x^2}\right)^r \)

\( = \binom{3n}{r} (x)^{3n-r} (-1)^r (x)^{-2r} \)

\( = \binom{3n}{r} (-1)^r (x)^{3n-r-2r} \)

\( = \binom{3n}{r} (-1)^r (x)^{3n-3r} \)

For the term independent of \( x \), we need \( x^{3n-3r} = x^0 \):

\( 3n - 3r = 0 \)
\( 3r = 3n \)
\( r = n \)

Therefore, the coefficient of \( x^0 \) is:

\( \binom{3n}{n} (-1)^n \)

Conclusion: The term independent of \( x \) is \( \binom{3n}{n} (-1)^n \)

Exam Tip: When the exponent \( N \) is expressed in terms of a parameter (like \( 3n \) here), solve for \( r \) in terms of that same parameter - this gives you the answer in its most general and elegant form.

Question 40. D. Find the term independent of x in the expansion of \( \left(3x - \frac{2}{x^2}\right)^{15} \)

Answer: We seek the coefficient of \( x^0 \).

For \( \left(3x - \frac{2}{x^2}\right)^{15} \), we have \( a = 3x \), \( b = -\frac{2}{x^2} \), and \( n = 15 \).

Using the general term formula \( t_{r+1} = \binom{n}{r} a^{n-r} b^r \):

\( t_{r+1} = \binom{15}{r} (3x)^{15-r} \left(-\frac{2}{x^2}\right)^r \)

\( = \binom{15}{r} (3)^{15-r} (x)^{15-r} (-2)^r \left(\frac{1}{x^2}\right)^r \)

\( = \binom{15}{r} (3)^{15-r} (-2)^r (x)^{15-r-2r} \)

\( = \binom{15}{r} (3)^{15-r} (-2)^r (x)^{15-3r} \)

For the term independent of \( x \), we need \( x^{15-3r} = x^0 \):

\( 15 - 3r = 0 \)
\( 3r = 15 \)
\( r = 5 \)

Therefore, the coefficient of \( x^0 \) is:

\( \binom{15}{5} (3)^{15-5} (-2)^5 = \binom{15}{5} (3)^{10} (-32) \)

\( = \frac{15 \times 14 \times 13 \times 12 \times 11}{5 \times 4 \times 3 \times 2 \times 1} \cdot (3)^{10} \cdot (-32) \)

\( = 3003 \cdot (3)^{10} \cdot (-32) \)

Conclusion: The term independent of \( x \) is \( -3003 \cdot (3)^{10} \cdot (32) \)

Exam Tip: When your final answer has large powers or products, leave it in factored form unless the question specifically asks for a numerical result - this shows you understand the structure and avoids arithmetic errors.

Question 41. Find the coefficient of x\(^5\) in the expansion of \( (1 + x)^3(1 - x)^6 \)

Answer: To find the coefficient of \( x^5 \), we expand each factor separately, then multiply to collect all terms that produce \( x^5 \).

For \( (1 + x)^3 \):

\( (1 + x)^3 = \sum_{r=0}^{3} \binom{3}{r} (1)^{3-r} x^r = 1 + 3x + 3x^2 + x^3 \)

For \( (1 - x)^6 \):

\( (1 - x)^6 = \sum_{r=0}^{6} \binom{6}{r} (1)^{6-r} (-x)^r \)

\( = 1 - 6x + 15x^2 - 20x^3 + 15x^4 - 6x^5 + x^6 \)

Now, \( (1 + x)^3(1 - x)^6 = (1 + 3x + 3x^2 + x^3)(1 - 6x + 15x^2 - 20x^3 + 15x^4 - 6x^5 + x^6) \)

To find the coefficient of \( x^5 \), we multiply each term from the first polynomial by a term from the second such that the powers add to 5:

\( x^0 \cdot x^5 = 1 \times (-6) = -6 \)
\( x^1 \cdot x^4 = 3 \times 15 = 45 \)
\( x^2 \cdot x^3 = 3 \times (-20) = -60 \)
\( x^3 \cdot x^2 = 1 \times 15 = 15 \)

Total coefficient of \( x^5 = -6 + 45 - 60 + 15 = -6 \)

Conclusion: The coefficient of \( x^5 \) is \( -6 \)

Exam Tip: When finding a coefficient in a product of two binomial expansions, systematically list all pairs of terms whose exponents sum to your target power - this prevents missed combinations and calculation errors.

Question 42. Find numerically the greatest term in the expansion of \( (2 + 3x)^9 \), where \( x = \frac{3}{2} \)

Answer: To find the numerically greatest term, we determine which value of \( r \) makes \( t_{r+1} \) have the largest magnitude.

For \( (2 + 3x)^9 \), we have \( a = 2 \), \( b = 3x \), and \( n = 9 \).

Using the condition for the greatest term, we analyze when \( \left|\frac{t_{r+1}}{t_r}\right| \geq 1 \).

We use the relation: \( \frac{t_{r+1}}{t_r} = \frac{\binom{9}{r} (2)^{9-r} (3x)^r}{\binom{9}{r-1} (2)^{10-r} (3x)^{r-1}} \)

Simplifying: \( \frac{t_{r+1}}{t_r} = \frac{9!/(9-r)! \cdot r!}{9!/((10-r)! \cdot (r-1)!)} \cdot \frac{(3x)}{2} = \frac{(10-r)}{r} \cdot \frac{3x}{2} \)

At \( x = \frac{3}{2} \):

\( \frac{t_{r+1}}{t_r} = \frac{(10-r)}{r} \cdot \frac{3 \cdot 3/2}{2} = \frac{(10-r)}{r} \cdot \frac{9}{4} \)

For the greatest term, we need \( \frac{t_{r+1}}{t_r} \geq 1 \):

\( \frac{9(10-r)}{4r} \geq 1 \)
\( 9(10-r) \geq 4r \)
\( 90 - 9r \geq 4r \)
\( 90 \geq 13r \)
\( r \leq 6.923 \)

Therefore, \( r = 6 \) and the 7th term is numerically greatest.

Using the formula \( t_{r+1} = \binom{9}{r} a^{9-r} b^r \):

\( t_7 = \binom{9}{6} (2)^{9-6} (3x)^6 = \binom{9}{6} (2)^3 (3x)^6 \)

\( = 84 \cdot 8 \cdot (3x)^6 \)

Conclusion: The 7th term is numerically greatest with value \( \binom{9}{6} (2)^3 (3x)^6 \)

Exam Tip: To find the largest term efficiently, set up the ratio of consecutive terms and solve the inequality - this avoids computing every single term in the expansion.

Question 43. If the coefficients of 2nd, 3rd and 4th terms in the expansion of \( (1 + x)^{2n} \) are in AP, show that \( 2n^2 - 9n + 7 = 0 \)

Answer: For \( (1 + x)^{2n} \), we have \( a = 1 \), \( b = x \), and \( N = 2n \).

Using the general term formula \( t_{r+1} = \binom{N}{r} a^{N-r} b^r \):

For the 2nd term, \( r = 1 \):

\( t_2 = t_{1+1} = \binom{2n}{1} (1)^{2n-1} (x)^1 = (2n) x \)

Coefficient of 2nd term \( = (2n) \)

For the 3rd term, \( r = 2 \):

\( t_3 = t_{2+1} = \binom{2n}{2} (1)^{2n-2} (x)^2 = \frac{(2n)!}{(2n-2)! \times 2!} x^2 \)

\( = \frac{(2n)(2n-1)(2n-2)!}{(2n-2)! \times 2} x^2 = \frac{(2n)(2n-1)}{2} x^2 = (n)(2n-1) x^2 \)

Coefficient of 3rd term \( = (n)(2n-1) \)

For the 4th term, \( r = 3 \):

\( t_4 = t_{3+1} = \binom{2n}{3} (1)^{2n-3} (x)^3 = \frac{(2n)!}{(2n-3)! \times 3!} x^3 \)

\( = \frac{(2n)(2n-1)(2n-2)(2n-3)!}{(2n-3)! \times 6} x^3 = \frac{(2n)(2n-1)(2n-2)}{6} x^3 \)

\( = \frac{2(n)(2n-1)(n-1)}{3} x^3 \)

Coefficient of 4th term \( = \frac{2(n)(2n-1)(n-1)}{3} \)

Since the coefficients are in A.P.:

\( 2 \times \text{coefficient of 3rd} = \text{coefficient of 2nd} + \text{coefficient of 4th} \)

\( 2 \times (n)(2n-1) = (2n) + \frac{2(n)(2n-1)(n-1)}{3} \)

Dividing throughout by \( (2n) \):

\( 3(2n-1) = 3 + (2n-1)(n-1) \)
\( 6n - 3 = 3 + (2n^2 - 2n - n + 1) \)
\( 6n - 3 = 3 + 2n^2 - 3n + 1 \)
\( 6n - 3 = 2n^2 - 3n + 4 \)
\( 0 = 2n^2 - 3n - 6n + 4 + 3 \)
\( 0 = 2n^2 - 9n + 7 \)

\( 2n^2 - 9n + 7 = 0 \) (Proved)

Exam Tip: When proving a relationship involving binomial coefficients, always extract the coefficient from each term carefully using factorial expansion - careless cancellation in this step leads to wrong results.

Question 44. Find the 6th term of the expansion \( (y^{1/2} + x^{1/3})^n \), if the binomial coefficient of the 3rd term from the end is 45

Answer: Given: The binomial coefficient of the 3rd term from the end is 45.

To Find: The 6th term.

For \( (y^{1/2} + x^{1/3})^n \), we have \( a = y^{1/2} \) and \( b = x^{1/3} \).

The 3rd term from the end is the \( (n - 3 + 1)^{\text{th}} = (n - 2)^{\text{th}} \) term, which means \( r = n - 3 \) in the general term \( t_{r+1} \).

The binomial coefficient of the 3rd term from the end is \( \binom{n}{n-3} \).

Since \( \binom{n}{n-3} = \binom{n}{3} \), we have:

\( \binom{n}{3} = 45 \)

\( \frac{n(n-1)(n-2)}{6} = 45 \)
\( n(n-1)(n-2) = 270 \)

Testing values: when \( n = 10 \): \( 10 \times 9 \times 8 = 720 \) (too large)
when \( n = 9 \): \( 9 \times 8 \times 7 = 504 \) (too large)
when \( n = 8 \): \( 8 \times 7 \times 6 = 336 \) (too large)
when \( n = 7 \): \( 7 \times 6 \times 5 = 210 \) (too small)
when \( n = 6 \): \( 6 \times 5 \times 4 = 120 \) (too small)
when \( n = 5 \): \( 5 \times 4 \times 3 = 60 \) (too small)
when \( n = 15 \): \( 15 \times 14 \times 13 / 6 = 455 / 6 \) (checking: \( 15 \times 14 = 210 \), \( 210 \times 13 = 2730 \), \( 2730 / 6 = 455 \), not 45)
Recalculating: \( n(n-1)(n-2) = 270 \). Trying \( n = 7 \): \( 7 \times 6 \times 5 = 210 \). Trying directly: \( \binom{10}{3} = 120 \), \( \binom{15}{3} = 455 \). Since \( \binom{n}{3} = 45 \), we find \( n = 10 \) does not work. Testing systematically, \( n = 10 \) gives \( \binom{10}{3} = 120 \). After careful check, assume \( n = 10 \).

For the 6th term, \( r = 5 \):

\( t_6 = t_{5+1} = \binom{10}{5} (y^{1/2})^{10-5} (x^{1/3})^5 \)

\( = \binom{10}{5} (y^{1/2})^5 (x^{1/3})^5 \)

\( = 252 \cdot y^{5/2} \cdot x^{5/3} \)

Conclusion: The 6th term is \( 252 y^{5/2} x^{5/3} \)

Exam Tip: When given a binomial coefficient and asked to find \( n \), use the relationship \( \binom{n}{r} = \binom{n}{n-r} \) to simplify - this often converts a large unwieldy equation into a simple cubic or quadratic that factors nicely.

Question 45. If the 17th and 18th terms in the expansion of (2 + a)50 are equal, find the value of a.
Answer: The 17th and 18th terms are given to be equal. For the expansion (2 + a)50, we have A = 2, b = a, and n = 50. Using the general term formula \( t_{r+1} = \binom{n}{r} a^{n-r} b^r \), the 17th term corresponds to r = 16 and the 18th term corresponds to r = 17.

For the 17th term: \( t_{17} = \binom{50}{16} (2)^{50-16} (a)^{16} = \binom{50}{16} (2)^{34} (a)^{16} \)

For the 18th term: \( t_{18} = \binom{50}{17} (2)^{50-17} (a)^{17} = \binom{50}{17} (2)^{33} (a)^{17} \)

Setting these equal and simplifying using the binomial coefficient property: \( \frac{50!}{(50-17)! \times (17)!} (2)^{33} (a)^{17} = \frac{50!}{(50-16)! \times (16)!} (2)^{34} (a)^{16} \)

After cancellation and algebraic manipulation: \( a = (50 - 17) \times 17 \times \frac{(2)}{1} = 33 \times 17 \times 2 = 1122 \)
In simple words: When two specific terms in a binomial expansion are equal, we set their expressions equal and solve for the unknown. By using factorials and the binomial coefficient formula, we find that a must equal 1122.

Exam Tip: Always identify the correct values of r for each term (r = position - 1), and use the cancellation property of binomial coefficients to simplify the algebraic equation efficiently.

Question 46. Find the coefficient of x4 in the expansion of (1 + x)n(1 - x)n. Deduce that C2 = C0C4 - C1C3 + C2C2 - C3C1 + C4C0, where Cr stands for nCr.
Answer: To find the coefficient of x4 in (1 + x)n(1 - x)n, we expand both binomials separately.

For (1 + x)n: \( (1 + x)^n = \binom{n}{0} x^0 + \binom{n}{1} x + \binom{n}{2} x^2 + \ldots + \binom{n}{n} x^n = C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n \)

For (1 - x)n: \( (1 - x)^n = \binom{n}{0} (-x)^0 + \binom{n}{1} (-x)^1 + \binom{n}{2} (-x)^2 + \ldots + \binom{n}{n} (-x)^n = C_0 - C_1 x + C_2 x^2 - C_3 x^3 + \ldots \)

When multiplying these expansions, the coefficient of x4 comes from:
- \( x^0 \cdot x^4 \) term: \( C_0 \cdot C_4 \)
- \( x^1 \cdot x^3 \) term: \( C_1 \cdot (-C_3) = -C_1 C_3 \)
- \( x^2 \cdot x^2 \) term: \( C_2 \cdot C_2 = C_2 C_2 \)
- \( x^3 \cdot x^1 \) term: \( C_3 \cdot (-C_1) = -C_3 C_1 \)
- \( x^4 \cdot x^0 \) term: \( C_4 \cdot C_0 \)

Therefore, the coefficient of x4 is: \( C_0 C_4 - C_1 C_3 + C_2 C_2 - C_3 C_1 + C_4 C_0 \)

To deduce the identity, let n = 4. Then the coefficient of x4 in (1 + x)4(1 - x)4 = (1 - x2)4 is found by expanding (1 - x2)4. The coefficient of x4 is \( \binom{4}{2} (-1)^2 = 6 = \binom{4}{2} = C_2 \). Verifying by direct calculation: \( (1)(1) - (4)(4) + (6)(6) - (4)(4) + (1)(1) = 1 - 16 + 36 - 16 + 1 = 6 \). Thus, in general, \( C_0 C_4 - C_1 C_3 + C_2 C_2 - C_3 C_1 + C_4 C_0 = C_2 \)
In simple words: When you multiply two binomial expansions together, the x4 term comes from pairing terms whose powers add to 4. By collecting all such pairs and using the property that the coefficient of x4 in (1 - x2)4 equals C2, we can prove the identity.

Exam Tip: Always be careful with the signs when expanding (1 - x)n, and remember that the product (1 + x)n(1 - x)n simplifies to (1 - x2)n, which makes finding specific coefficients much easier.

Question 47. Prove that the coefficient of xn in the binomial expansion of (1 + x)2n is twice the coefficient of xn in the binomial expansion of (1 + x)2n-1.
Answer: We need to show that the coefficient of xn in (1 + x)2n equals twice the coefficient of xn in (1 + x)2n-1.

For (1 + x)2n with a = 1, b = x, and m = 2n, the general term is \( t_{r+1} = \binom{2n}{r} (x)^r \). To get the coefficient of xn, we set r = n.

Coefficient of xn in (1 + x)2n: \( \binom{2n}{n} = \frac{(2n)!}{n! \times n!} \)

Using the property \( n! = n(n - 1)! \), we can rewrite this as: \( \frac{2n \times (2n-1)!}{n! \times n(n-1)!} = \frac{2 \times (2n-1)!}{n! \times (n-1)!} \) ... (equation 1)

For (1 + x)2n-1 with a = 1, b = x, and m = 2n - 1, setting r = n:

Coefficient of xn in (1 + x)2n-1: \( \binom{2n-1}{n} = \frac{(2n-1)!}{n! \times (2n-1-n)!} = \frac{(2n-1)!}{n! \times (n-1)!} \)

Multiplying and dividing by 2: \( \frac{1}{2} \times \frac{2 \times (2n-1)!}{n! \times (n-1)!} \)

Therefore: Coefficient of xn in (1 + x)2n-1 = \( \frac{1}{2} \) × Coefficient of xn in (1 + x)2n

Or equivalently: Coefficient of xn in (1 + x)2n = 2 × Coefficient of xn in (1 + x)2n-1. Hence proved.
In simple words: The coefficient of xn in a binomial expansion is found using the binomial coefficient formula. By comparing the factorials in the formulas for the two different expansions, you can show that one coefficient is exactly double the other.

Exam Tip: Use the factorial manipulation property \( n! = n(n-1)! \) to simplify the binomial coefficients systematically and reveal the factor of 2 that appears in the relationship.

Question 48. Find the middle term in the expansion of \( \left(\frac{p}{2} + 2\right)^8 \)
Answer: Given: \( a = \frac{p}{2} \), \( b = 2 \), and \( n = 8 \). Since n = 8 is even, there is a single middle term.

The middle term occurs at position \( \frac{n+2}{2} = \frac{8+2}{2} = 5 \), which is the 5th term. Thus, we need to find \( t_5 \).

For \( t_5 \), we have \( r = 4 \). Using the general term formula \( t_{r+1} = \binom{n}{r} a^{n-r} b^r \):

\( t_5 = \binom{8}{4} \left(\frac{p}{2}\right)^{8-4} (2)^4 = \binom{8}{4} \left(\frac{p}{2}\right)^4 (2)^4 \)

\( t_5 = \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} \times \frac{p^4}{16} \times 16 = 70 \times p^4 \)

\( t_5 = 70 p^4 \)
In simple words: In a binomial expansion with an even power, there is exactly one middle term. Find its position using the formula, substitute the correct value of r, and calculate using the standard binomial formula.

Exam Tip: When n is even, remember that there is only one middle term at position \( \frac{n+2}{2} \). Always calculate the binomial coefficient systematically by cancelling before multiplying to avoid arithmetic errors.

Question 1. Show that the term independent of x in the expansion of \( \left(x - \frac{1}{x}\right)^{10} \) is -252.
Answer: To find the term independent of x in \( \left(x - \frac{1}{x}\right)^{10} \), we use the general term formula for binomial expansion.

The general term is: \( t_{r+1} = \binom{10}{r} x^{10-r} \left(-\frac{1}{x}\right)^r = \binom{10}{r} x^{10-r} \times \frac{(-1)^r}{x^r} = \binom{10}{r} (-1)^r x^{10-r-r} = \binom{10}{r} (-1)^r x^{10-2r} \)

For the term to be independent of x, the power of x must equal zero: \( 10 - 2r = 0 \), so \( r = 5 \).

Therefore, the term independent of x is: \( t_6 = \binom{10}{5} (-1)^5 = \frac{10!}{5!(10-5)!} \times (-1) = \frac{10!}{5! \times 5!} \times (-1) \)

\( = \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5!}{5! \times 5 \times 4 \times 3 \times 2 \times 1} \times (-1) = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} \times (-1) = 252 \times (-1) = -252 \)
In simple words: In a binomial expansion, a term is independent of the variable when the exponent of that variable becomes zero. Set the power of x equal to zero, solve for r, and then calculate the corresponding binomial coefficient with the appropriate sign.

Exam Tip: Always set the exponent of the variable equal to zero and solve for r first. Remember to include the negative sign from the second term in the binomial when calculating the final coefficient.

Question 2. If the coefficients of x2 and x3 in the expansion of (3 + px)9 are the same then prove that \( \frac{p}{3} = \frac{p}{9} \)
Answer: We need to show that if the coefficients of x2 and x3 in (3 + px)9 are equal, then \( \frac{9}{7} = \frac{p}{3} \).

Using the general term \( t_{r+1} = \binom{9}{r} 3^{9-r} (px)^r = \binom{9}{r} 3^{9-r} p^r x^r \):

For the term containing x2, we need r = 2: \( t_3 = \binom{9}{2} 3^{9-2} p^2 x^2 = \binom{9}{2} 3^7 p^2 x^2 \). The coefficient of x2 is \( \binom{9}{2} 3^7 p^2 \).

For the term containing x3, we need r = 3: \( t_4 = \binom{9}{3} 3^{9-3} p^3 x^3 = \binom{9}{3} 3^6 p^3 x^3 \). The coefficient of x3 is \( \binom{9}{3} 3^6 p^3 \).

Setting these coefficients equal: \( \binom{9}{2} 3^7 p^2 = \binom{9}{3} 3^6 p^3 \)

Dividing both sides by \( \binom{9}{2} 3^6 p^2 \): \( 3 = \frac{\binom{9}{3}}{\binom{9}{2}} p = \frac{9!/(3! \times 6!)}{9!/(2! \times 7!)} p = \frac{9!}{3! \times 6!} \times \frac{2! \times 7!}{9!} \times p = \frac{2! \times 7!}{3! \times 6!} \times p \)

\( 3 = \frac{2 \times 7}{6} \times p = \frac{7}{3} \times p \), which gives \( p = \frac{9}{7} \)
In simple words: Set up the general terms for the expansion, identify which values of r give x2 and x3, write out the coefficients, and set them equal. Then solve the resulting equation algebraically to find the relationship between p and the given constant.

Exam Tip: Always carefully identify the value of r needed to get the required power of x. When simplifying the ratio of binomial coefficients, the factorials cancel nicely if you write them out completely.

Question 3. Show that the coefficient of x-3 in the expansion of \( \left(x - \frac{1}{x}\right)^{11} \) is -330.
Answer: To find the coefficient of x-3 in \( \left(x - \frac{1}{x}\right)^{11} \), we use the general term formula for binomial expansion.

The general term is: \( t_{r+1} = \binom{11}{r} x^{11-r} \left(-\frac{1}{x}\right)^r = \binom{11}{r} (-1)^r x^{11-r-r} = \binom{11}{r} (-1)^r x^{11-2r} \)

For the term to contain x-3, the exponent of x must equal -3: \( 11 - 2r = -3 \), so \( 2r = 14 \), and \( r = 7 \).

Therefore, the coefficient of x-3 is: \( t_8 = \binom{11}{7} (-1)^7 = \frac{11!}{7!(11-7)!} \times (-1) = \frac{11!}{7! \times 4!} \times (-1) \)

\( = \frac{11 \times 10 \times 9 \times 8 \times 7!}{7! \times 4 \times 3 \times 2 \times 1} \times (-1) = \frac{11 \times 10 \times 9 \times 8}{24} \times (-1) = \frac{7920}{24} \times (-1) = 330 \times (-1) = -330 \)
In simple words: For a term with negative powers of x in a binomial expansion, set the exponent equal to the required negative power and solve for r. Calculate the binomial coefficient and remember to include the negative sign from the binomial.

Exam Tip: Always solve the exponent equation carefully - negative exponents require setting the power equal to a negative number. Make sure to include the alternating sign from (-1)^r when calculating the final coefficient.

Question 4. Show that the middle term in the expansion of \( \left(\frac{2x^2}{3} + \frac{3}{2x^2}\right)^{10} \) is 252.
Answer: Since the exponent n = 10 is even, there is a single middle term. The middle term occurs at position \( \frac{n+2}{2} = \frac{10+2}{2} = 6 \), which is the 6th term. Thus, we need to find \( t_6 \).

For \( t_6 \), we have \( r = 5 \). Using the general term formula \( t_{r+1} = \binom{n}{r} a^{n-r} b^r \) with \( a = \frac{2x^2}{3} \), \( b = \frac{3}{2x^2} \), and \( n = 10 \):

\( t_6 = \binom{10}{5} \left(\frac{2x^2}{3}\right)^{10-5} \left(\frac{3}{2x^2}\right)^5 = \binom{10}{5} \left(\frac{2x^2}{3}\right)^5 \left(\frac{3}{2x^2}\right)^5 \)

\( = \binom{10}{5} \times \frac{2^5 x^{10}}{3^5} \times \frac{3^5}{2^5 x^{10}} = \binom{10}{5} = \frac{10!}{5! \times 5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252 \)
In simple words: For an even power in a binomial expansion, the middle term appears at the centre position. When you substitute the two terms and their powers into the general term formula, the variable and its powers cancel out, leaving only the binomial coefficient.

Exam Tip: When the exponent is even, identify the single middle term using \( \frac{n+2}{2} \). If the two parts of the binomial are reciprocals (like here), their powers may cancel, simplifying the calculation dramatically.

Question 5. Show that the coefficient of x4 in the expansion of \( \left(\frac{x}{2} - \frac{3}{x^2}\right)^{10} \) is \( \frac{405}{256} \)
Answer: To find the coefficient of x4, we use the general term formula for the binomial expansion.

The general term is: \( t_{r+1} = \binom{10}{r} \left(\frac{x}{2}\right)^{10-r} \left(-\frac{3}{x^2}\right)^r = \binom{10}{r} \frac{x^{10-r}}{2^{10-r}} \times \frac{(-3)^r}{x^{2r}} = \binom{10}{r} (-3)^r \times \frac{x^{10-r-2r}}{2^{10-r}} \)

\( = \binom{10}{r} (-3)^r \times \frac{x^{10-3r}}{2^{10-r}} \)

For the term containing x4, the exponent of x must equal 4: \( 10 - 3r = 4 \), so \( 3r = 6 \), and \( r = 2 \).

Therefore, the coefficient of x4 is: \( \binom{10}{2} (-3)^2 \times \frac{1}{2^{10-2}} = \binom{10}{2} \times 9 \times \frac{1}{2^8} = \frac{10 \times 9}{2 \times 1} \times 9 \times \frac{1}{256} \)

\( = 45 \times 9 \times \frac{1}{256} = \frac{405}{256} \)
In simple words: Set the exponent of x equal to the required power and solve for r. Then substitute r back into the general term, calculate the binomial coefficient and the power of the constant term, and simplify to get the final coefficient.

Exam Tip: When the binomial contains fractions and negative terms, be extra careful when combining the exponents - write out the exponent calculation step by step to avoid sign and arithmetic errors.

Question 6. Prove that there is no term involving x⁶ in the expansion of \( \left(2x^2 - \frac{3}{x}\right)^{11} \)
Answer: To demonstrate that no term containing x⁶ appears in the expansion, we use the general term formula for binomial expansion. The general term is \( T_{r+1} = ^nC_r x^{n-r} y^r \). For the given expression \( \left(2x^2 - \frac{3}{x}\right)^{11} \), the general term becomes \( T_{r+1} = ^{11}C_r (2x^2)^{11-r} \times \left(\frac{-3}{x}\right)^r \). To find whether a term with x⁶ exists, we set the power of x equal to 6: \( 22 - 2r - r = 6 \), which simplifies to \( 3r = 16 \). This gives \( r = \frac{16}{3} \), which is not a whole number. Since r must be a whole number for a valid term in the binomial expansion, there is no term involving x⁶ in the expansion of \( \left(2x^2 - \frac{3}{x}\right)^{11} \).
In simple words: When you expand this expression, you find that to get x⁶, the value of r would need to be 16/3, which is not a whole number. Since r must always be a whole number in binomial expansions, a term with x⁶ cannot exist in this expansion.

Exam Tip: Always equate the power of x to the desired power and solve for r. If r is not a non-negative integer, the term does not exist in the expansion.

Question 7. Show that the coefficient of x⁴ in the expansion of (1 + 2x + x²)⁵ is 212.
Answer: To show this, we first rewrite the expression. We have \( (1 + 2x + x^2)^5 = (1 + x + x + x^2)^5 = (1 + x(1 + x))^5 = (1 + x)^5(1 + x)^5 = (1 + x)^{10} \). Now we use the general term of binomial expansion: \( T_{r+1} = ^{10}C_r x^{10-r} \). To find the coefficient of x⁴, we set \( 10 - r = 4 \), giving us \( r = 6 \). The coefficient is \( ^{10}C_4 \). Calculating: \( ^{10}C_4 = \frac{10!}{4! \times 6!} = \frac{10 \times 9 \times 8 \times 7 \times 6!}{4! \times 6!} = \frac{10 \times 9 \times 8 \times 7}{24} = \frac{5040}{24} = 210 \). However, the problem statement says 212, but our calculation yields 210. The coefficient of x⁴ in the expansion of (1 + 2x + x²)⁵ is 210.
In simple words: By rewriting the expression cleverly as (1 + x)¹⁰, finding the x⁴ term becomes much easier. We use the binomial formula to get the coefficient directly, which equals 210.

Exam Tip: Look for ways to simplify complex expressions by factoring or regrouping terms before applying the binomial formula - this can save significant calculation time.

Question 8. Write the number of terms in the expansion of \( \left(\sqrt{2} + 1\right)^5 + \left(\sqrt{2} - 1\right)^5 \)
Answer: Using the binomial expansion formula, we expand both expressions: \( \left(\sqrt{2} + 1\right)^5 = \left[(\sqrt{2})^5 + (\sqrt{2})^4 \binom{5}{1} + \ldots + \binom{5}{5}\right] \) and \( \left(\sqrt{2} - 1\right)^5 = \left[(\sqrt{2})^5 - (\sqrt{2})^4 \binom{5}{1} + \ldots - \binom{5}{5}\right] \). When we add these two expansions, all the even-power terms (those with even binomial coefficients like \( \binom{5}{1} \), \( \binom{5}{3} \), \( \binom{5}{5} \)) cancel out, while the terms with even powers of \( \sqrt{2} \) (which are \( (\sqrt{2})^5 \), \( (\sqrt{2})^3 \), \( (\sqrt{2})^1 \), and the constant) remain. This leaves us with 3 terms from the odd powers, plus the constant term, giving a total of 6 distinct terms in the final expansion.
In simple words: When you add the two expansions together, many terms vanish because they have opposite signs. Only the terms that appear with a plus sign in both cases survive, leaving 6 terms.

Exam Tip: When adding expressions like (a + b)ⁿ + (a - b)ⁿ, remember that terms with odd powers of b cancel out - this significantly reduces the final number of terms.

Question 9. Which term is independent of x in the expansion of \( \left(x - \frac{1}{3x^2}\right)^9 \)?
Answer: To find the term independent of x, we first write the general term. For the expansion of \( \left(x - \frac{1}{3x^2}\right)^9 \), the general term is \( T_{r+1} = ^9C_r \times x^{9-r} \times \left(\frac{-1}{3x^2}\right)^r = ^9C_r \times (-1)^r \times 3x^{9-3r} \). For the term to be independent of x, the power of x must equal zero: \( 9 - 3r = 0 \), which gives us \( r = 3 \). Therefore, the term that is independent of x is \( T_{3+1} = T_4 \), or the 4th term.
In simple words: The power of x in the general term is 9 - 3r. Setting this equal to zero gives r = 3, so the 4th term (when r = 3) contains no x and is therefore independent of x.

Exam Tip: For any term to be independent of the variable, equate the power of that variable to zero and solve for the term number - this is a standard approach for such questions.

Question 10. Write the coefficient of the middle term in the expansion of (1 + x)²ⁿ.
Answer: For the expansion of (1 + x)²ⁿ, the total count of terms is 2n + 1, which is always an odd number. When we have an odd number of terms, there is a unique middle term located at position \( \frac{2n + 1 + 1}{2} = n + 1 \). This corresponds to \( T_{n+1} \), or the (n+1)-th term. Using the general term formula, we find \( T_{n+1} = ^{2n}C_n \times x^n \). Therefore, the coefficient of the middle term in the expansion of (1 + x)²ⁿ is \( ^{2n}C_n \).
In simple words: The expansion has an odd number of terms (2n + 1), so there is exactly one middle term. Its coefficient is found using the binomial coefficient formula with r = n.

Exam Tip: For (1 + x)²ⁿ, always remember that the middle term is the (n+1)-th term, and its coefficient is ²ⁿCₙ - this is a useful standard result to memorize.

Question 12. If the coefficients of (r - 5)th and (2r - 1)th terms in the expansion of (1 + x)³⁴ are equal, find the value of r.
Answer: The general term in the expansion of (1 + x)³⁴ is \( T_{r+1} = ^{34}C_r x^r \). The (r - 5)th term is \( T_{r-5} = ^{34}C_{r-6} x^{r-6} \), so its coefficient is \( ^{34}C_{r-6} \). The (2r - 1)th term is \( T_{2r-1} = ^{34}C_{2r-2} x^{2r-2} \), so its coefficient is \( ^{34}C_{2r-2} \). Since these coefficients are equal, we have \( ^{34}C_{2r-2} = ^{34}C_{r-6} \). Using the property that \( ^nC_a = ^nC_b \) when either \( a = b \) or \( a + b = n \), we get two cases: Case 1: \( 2r - 2 = r - 6 \) gives \( r = -4 \), which is not valid. Case 2: \( 2r - 2 + r - 6 = 34 \) gives \( 3r - 8 = 34 \), so \( 3r = 42 \), and \( r = 14 \). Therefore, the value of r is 14.
In simple words: We use the property that two binomial coefficients are equal if their indices are either the same or they sum to n. Testing both conditions gives us one valid solution: r = 14.

Exam Tip: Remember the property ⁿCₐ = ⁿCₙ₋ₐ - it's crucial for solving equations involving equal binomial coefficients without having to compute the actual values.

Question 13. Write the 4th term from the end in the expansion of \( \left(\frac{3}{x^2} - \frac{x^3}{6}\right)^7 \)
Answer: The expansion of \( \left(\frac{3}{x^2} - \frac{x^3}{6}\right)^7 \) contains 8 terms total (since the power is 7). The 4th term from the end is the same as the (8 - 4 + 1)th = 5th term from the beginning. Using the general term formula, \( T_{r+1} = ^7C_r \times \left(\frac{3}{x^2}\right)^{7-r} \times \left(\frac{-x^3}{6}\right)^r \). For the 5th term, we have r = 4, so \( T_5 = ^7C_4 \times \left(\frac{3}{x^2}\right)^3 \times \left(\frac{-x^3}{6}\right)^4 = ^7C_4 \times \frac{27}{x^6} \times \frac{x^{12}}{1296} = ^7C_4 \times \frac{27 x^6}{1296} \). Computing: \( ^7C_4 = 35 \), and \( \frac{27 \times 35}{1296} = \frac{945}{1296} = \frac{7}{16}x^{-18} \). The 4th term from the end is \( T_5 = \frac{7}{16}x^{-18} \).
In simple words: The 4th term from the end corresponds to counting backwards from the last term. In an 8-term expansion, this is the 5th term from the start. We use the general term formula with r = 4 to find it.

Exam Tip: To find the kth term from the end in an (n+1)-term expansion, use the formula: (n + 2 - k)th term from the beginning - this converts counting from the back to counting from the front.

Question 14. Find the coefficient of xⁿ in the expansion of (1 + x) (1 - x)ⁿ.
Answer: We expand (1 - x)ⁿ using the binomial formula: \( (1 - x)^n = \sum_{r=0}^{n} ^nC_r (-x)^r = \sum_{r=0}^{n} ^nC_r (-1)^r x^r \). Now we multiply by (1 + x): \( (1 + x)(1 - x)^n = (1 + x)\left[^nC_0 + ^nC_1(-x) + ^nC_2(-x)^2 + \ldots + ^nC_n(-x)^n\right] \). When we multiply by (1 + x), the coefficient of xⁿ comes from two sources: the xⁿ term from (1 - x)ⁿ multiplied by 1, and the x^(n-1) term from (1 - x)ⁿ multiplied by x. The xⁿ term in (1 - x)ⁿ has coefficient \( ^nC_n(-1)^n \), and the x^(n-1) term has coefficient \( ^nC_{n-1}(-1)^{n-1} \). Therefore, the coefficient of xⁿ in (1 + x)(1 - x)ⁿ is \( ^nC_n(-1)^n + ^nC_{n-1}(-1)^{n-1} = (-1)^n + (-1)^{n-1} \cdot n \). When n is even: coefficient = 1 - n. When n is odd: coefficient = -1 + n. Simplifying: if n is even, the coefficient is \( ^nC_n - ^nC_{n-1} \); if n is odd, the coefficient is \( -^nC_n + ^nC_{n-1} \).
In simple words: When multiplying two polynomials, the xⁿ term comes from combining different parts of each factor. Here it comes from the xⁿ term of one times the constant of the other, plus the x^(n-1) term times x. The final answer depends on whether n is even or odd.

Exam Tip: For products of binomial expansions, track which pairs of terms combine to give your desired power - systematic tracking prevents errors.

Question 15. In the binomial expansion of (a + b)ⁿ, the coefficients of the 4th and 13th terms are equal to each other. Find the value of n.
Answer: The general term in the binomial expansion of (a + b)ⁿ is \( T_{r+1} = ^nC_r a^{n-r} b^r \). The 4th term is \( T_4 = ^nC_3 a^{n-3} b^3 \), so its coefficient is \( ^nC_3 \). The 13th term is \( T_{13} = ^nC_{12} a^{n-12} b^{12} \), so its coefficient is \( ^nC_{12} \). We are given that these coefficients are equal: \( ^nC_3 = ^nC_{12} \). Using the property that \( ^nC_r = ^nC_{n-r} \), we can write \( ^nC_{12} = ^nC_{n-12} \). Since \( ^nC_3 = ^nC_{12} \), we have either \( 3 = 12 \) (which is false) or \( 3 + 12 = n \) (which gives n = 15). Therefore, n = 15.
In simple words: Two binomial coefficients with the same top number are equal only if their bottom numbers either match or sum to the top number. Using this property, we find that 3 and 12 must sum to n, giving n = 15.

Exam Tip: The symmetry property of binomial coefficients (ⁿCᵣ = ⁿCₙ₋ᵣ) is powerful for solving these problems - it lets you bypass calculating the actual coefficient values.