ML Aggarwal Class 11 Maths Solutions Chapter 08 Permutations and Combinations

Binomial Theorem

Introduction

You have already learned about the squares and cubes of binomial expressions such as a + b, a - b. You have used these to find the values of numbers like (103)² and (998)³ by writing them as (103)² = (100 + 3)² and (998)³ = (1000 - 2)³. However, when dealing with higher powers like (103)⁷ or (998)⁹, repeated multiplication becomes very tedious. A result called the Binomial theorem solves this problem. The general form of the binomial expression is a + b, and the expansion of (a + b)ⁿ, where n ∈ N, is called the binomial theorem for positive integral index. The binomial theorem allows us to expand any power of a binomial expression. Sir Isaac Newton first discovered it.

Development of Binomial Theorem

We know that

\( (a + b)^0 = 1 \) (Assume a + b ≠ 0)
\( (a + b)^1 = a + b \)
\( (a + b)^2 = a^2 + 2ab + b^2 \)
\( (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \)
\( (a + b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4 \) etc.

From the above expansions, we observe that:

(i) Each expansion has a total number of terms that is one more than its index. For example, the expansion of (a + b)³ contains 4 terms, while the index of (a + b)³ is 3.

(ii) The power (index) of the first quantity 'a' decreases by 1 with each successive term, whereas the power of the second quantity 'b' increases by 1.

(iii) In each expansion, the sum of the indices of a and b remains the same and equals the index of (a + b). For example, in every term of (a + b)³, the sum of the indices of a and b is 3.

The coefficients of the terms in the above expansions follow a pattern and can be displayed in a table as follows:

These coefficients follow a specific pattern. We observe that:

(i) Each row starts with 1 and ends with 1.

(ii) Starting from the third row onwards, each coefficient (except the first and last) equals the sum of the two coefficients in the preceding row - one directly above it and one to its upper right.

This pattern is known as Pascal's Triangle.

In this arrangement, the numbers involved in addition and their results can be shown as displayed in the following table. This table may be extended by adding additional rows:

The table can be continued to any index we choose. Using Pascal's triangle, we can write expansions for higher powers of binomial expressions. For example, to expand (a + b)⁶ using Pascal's triangle, we use the row for index 6, which is:

1   6   15   20   15   6   1

Using these coefficients along with observations (i), (ii), and (iii), we get:

\( (a + b)^6 = a^6 + 6a^5b + 15a^4b^2 + 20a^3b^3 + 15a^2b^4 + 6ab^5 + b^6 \)

By using the concept of combinations, where \( ^nC_r = \frac{n!}{(n-r)! \cdot r!} \), with \( 0 \leq r \leq n \) and n a non-negative integer, and also \( ^nC_n = 1 = ^nC_0 \), the binomial expansions can be written as:

\( (a + b)^1 = a + b = ^1C_0 a^1 + ^1C_1 b^1 \)
\( (a + b)^2 = a^2 + 2ab + b^2 = ^2C_0 a^2 + ^2C_1 a^{2-1} b^1 + ^2C_2 b^2 \)
\( (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 = ^3C_0 a^3 + ^3C_1 a^{3-1} b^1 + ^3C_2 a^{3-2} b^2 + ^3C_3 b^3 \) etc.

By examining the above expansions, we can readily identify the general formula for the expansion of (a + b)ⁿ, where n ∈ N.

8.1 Binomial Theorem for Positive Integral Index

If n is a natural number and a and b are any numbers, then:

\( (a + b)^n = ^nC_0a^n + ^nC_1a^{n-1}b + ^nC_2a^{n-2}b^2 + \ldots + ^nC_{n-1}ab^{n-1} + ^nC_nb^n \)

Proof. We shall prove this theorem using the principle of mathematical induction.

Let P(n) be the statement:

\( (a + b)^n = ^nC_0 a^n + ^nC_1 a^{n-1}b + ^nC_2 a^{n-2}b^2 + \ldots + ^nC_{n-1}ab^{n-1} + ^nC_n b^n \)

For P(1), we have:

\( (a + b)^1 = ^1C_0 a^1 + ^1C_1b^1 \)
i.e., \( a + b = 1 \times a + 1 \times b \), which is true
\( \implies P(1) \) is true.

Assume P(m) is true:

\( (a + b)^m = ^mC_0a^m + ^mC_1a^{m-1}b + ^mC_2a^{m-2}b^2 + \ldots + ^mC_{m-1}ab^{m-1} + ^mC_m b^m \) ...(i)

For P(m + 1):

\( (a + b)^{m+1} = (a + b)^m (a + b) \)
\( = (^mC_0 a^m + ^mC_1a^{m-1}b + ^mC_2 a^{m-2}b^2 + \ldots + ^mC_{m-1}ab^{m-1} + ^mC_mb^m) (a + b) \) (using (i))
\( = ^mC_0 a^{m+1} + ^mC_1a^mb + ^mC_2a^{m-1}b^2 + \ldots + ^mC_{m-1}a^2b^{m-1} + ^mC_mab^m + ^mC_0a^mb + ^mC_1 a^{m-1}b^2 + ^mC_2a^{m-2}b^3 + \ldots + ^mC_{m-1}ab^m + ^mC_m b^{m+1} \) (by actual multiplication)
\( = ^mC_0a^{m+1} + (^mC_1 + ^mC_0) a^mb + (^mC_2 + ^mC_1) a^{m-1} b^2 + \ldots + (^mC_m + ^mC_{m-1}) ab^m + ^mC_mb^{m+1} \) (grouping like terms)
\( = ^{m+1}C_0 a^{m+1} + ^{m+1}C_1 a^mb + ^{m+1}C_2 a^{m-1} b^2 + \ldots + ^{m+1}C_m ab^m + ^{m+1}C_{m+1} b^{m+1} \)

This is true because we know that \( ^mC_0 = 1 = ^{m+1}C_0 \), \( ^mC_m = 1 = ^{m+1}C_{m+1} \), and \( ^mC_r + ^mC_{r-1} = ^{m+1}C_r \) for \( r = 1, 2, 3, \ldots, m \).

Therefore, \( ^mC_1 + ^mC_0 = ^{m+1}C_1 \), \( ^mC_2 + ^mC_1 = ^{m+1}C_2 \), ..., \( ^mC_m + ^mC_{m-1} = ^{m+1}C_m \)

\( \implies P(m + 1) \) is true.

Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

The notation \( \sum_{r=0}^{n} {^nC_r a^{n-r} b^r} \) stands for:

\( ^nC_0 a^nb^0 + ^nC_1 a^{n-1}b + ^nC_2 a^{n-2} b^2 + \ldots + ^nC_r a^{n-r} b^r + \ldots + ^nC_n a^{n-n} b^n \)

(Note that \( b^0 = 1 \) and \( a^{n-n} = a^0 = 1 \))

Therefore, the binomial theorem can be written as:

\( (a + b)^n = \sum_{r=0}^{n} {^nC_r a^{n-r} b^r} \)

8.1.1 Some Important Observations

1. The total number of terms in the expansion of (a + b)ⁿ is (n + 1), which is one more than the index n.

2. The sum of indices of a and b in each term is n. In the first term of (a + b)ⁿ, the index of a starts at n and decreases by 1 in every successive term until it reaches 0. Meanwhile, the index of b starts at 0 and increases by 1 in every successive term until it reaches n.

3. The coefficients \( ^nC_0, ^nC_1, ^nC_2, \ldots, ^nC_n \) are called binomial coefficients.

4. Since \( ^nC_r = ^nC_{n-r} \) for \( r = 0, 1, 2, \ldots, n \), we have \( ^nC_0 = ^nC_n \), \( ^nC_1 = ^nC_{n-1} \), \( ^nC_2 = ^nC_{n-2} \), ... Therefore, the coefficients of terms that are equidistant from the beginning and end are equal.

8.1.2 Some Special Cases

1. By replacing 'b' with '- b' in the binomial expansion of (a + b)ⁿ, we get:

\( (a - b)^n = ^nC_0 a^n + ^nC_1 a^{n-1}(- b) + ^nC_2 a^{n-2}(- b)^2 + \ldots + ^nC_r a^{n-r}(- b)^r + \ldots + ^nC_n(- b)^n \)
\( = ^nC_0 a^n - ^nC_1 a^{n-1} b + ^nC_2 a^{n-2} b^2 + \ldots + (- 1)^r {^nC_r a^{n-r} b^r} + \ldots + (- 1)^n {^nC_n b^n} \)
\( = \sum_{r=0}^{n} {(- 1)^r {^nC_r a^{n-r} b^r}} \)

As a result, the terms in the expansion of (a - b)ⁿ alternate between positive and negative. The final term is positive when n is even and negative when n is odd.

2. By putting a = 1 and b = x in the binomial expansion of (a + b)ⁿ, we get:

\( (1 + x)^n = ^nC_0 1^n + ^nC_1 1^{n-1} x + ^nC_2 1^{n-2} x^2 + \ldots + ^nC_r 1^{n-r} x^r + \ldots + ^nC_n x^n \)
\( = ^nC_0 + ^nC_1 x + ^nC_2 x^2 + \ldots + ^nC_r x^r + \ldots + ^nC_n x^n \)
\( = \sum_{r=0}^{n} {^nC_r x^r} \)

3. By putting a = 1 and b = - x in the binomial expansion of (a + b)ⁿ, we get:

\( (1 - x)^n = ^nC_0 - ^nC_1 x + ^nC_2 x^2 + \ldots + (- 1)^r {^nC_r x^r} + \ldots + (- 1)^n {^nC_n x^n} \)
\( = \sum_{r=0}^{n} {(- 1)^r {^nC_r x^r}} \)

4. In the expansion of (1 + x)ⁿ, where n ∈ N:

(i) \( ^nC_0 + ^nC_1 + ^nC_2 + \ldots + ^nC_r + \ldots + ^nC_n = 2^n \)

(ii) \( ^nC_0 - ^nC_1 + ^nC_2 - ^nC_3 + \ldots + (- 1)^n {^nC_n} = 0 \)

(iii) \( ^nC_0 + ^nC_2 + ^nC_4 + \ldots = ^nC_1 + ^nC_3 + ^nC_5 + \ldots = 2^{n-1} \)

Proof. We know that:

\( (1 + x)^n = ^nC_0 + ^nC_1 x + ^nC_2 x^2 + \ldots + ^nC_r x^r + \ldots + ^nC_n x^n \) ...(1)

(i) By substituting x = 1 in (1), we obtain:

\( (1 + 1)^n = ^nC_0 + ^nC_1 . 1 + ^nC_2 . 1^2 + \ldots + ^nC_r . 1^r + \ldots + ^nC_n . 1^n \)
\( \implies ^nC_0 + ^nC_1 + ^nC_2 + \ldots + ^nC_r + \ldots + ^nC_n = 2^n \)

Therefore, the sum of the binomial coefficients in the expansion of (1 + x)ⁿ, where n ∈ N, is 2ⁿ.

(ii) By substituting x = - 1 in (1), we obtain:

\( (1 - 1)^n = ^nC_0 + ^nC_1 (- 1) + ^nC_2 (- 1)^2 + ^nC_3 (- 1)^3 + \ldots + ^nC_n (- 1)^n \)
\( \implies ^nC_0 - ^nC_1 + ^nC_2 - ^nC_3 + \ldots + (- 1)^n {^nC_n} = 0 \)

(iii) From part (ii), we get:

\( ^nC_0 + ^nC_2 + ^nC_4 + \ldots = ^nC_1 + ^nC_3 + ^nC_5 + \ldots \) Therefore, the sum of each equals \( \frac{1}{2} \) (sum of the coefficients of all terms)
\( = \frac{1}{2} . 2^n \) (using part (i))
\( = 2^{n-1} \)
\( \implies ^nC_0 + ^nC_2 + ^nC_4 + \ldots = ^nC_1 + ^nC_3 + ^nC_5 + \ldots = 2^{n-1} \)

As a result, in the expansion of (1 + x)ⁿ where n ∈ N, the sum of the coefficients of odd-positioned terms equals the sum of the coefficients of even-positioned terms, and both equal 2^(n-1).

Remarks

1. If n is a positive odd integer, then both (a + b)ⁿ + (a - b)ⁿ and (a + b)ⁿ - (a - b)ⁿ contain the same number of terms, which equals \( \frac{n+1}{2} \).

2. If n is a positive even integer, then:

(i) (a + b)ⁿ + (a - b)ⁿ has \( \frac{n}{2} + 1 \) terms, and

(ii) (a + b)ⁿ - (a - b)ⁿ has \( \frac{n}{2} \) terms.

Question 1. Find the number of terms in the expansions of the following:
(i) \( (7x + 2y)^9 \)
(ii) \( \left(2x - \frac{3}{x^3}\right)^{10} \)
(iii) \( (1 + 2x + x^2)^{11} \)
(iv) \( (x + 2y - 3z)^n, n \in N \)
Answer: Since the number of terms in the expansion of (x + a)ⁿ is (n + 1), the number of terms in the expansion of (7x + 2y)⁹ is 9 + 1 = 10. For the second expression, the total number of terms is 10 + 1 = 11. The third expansion equals (1 + 2x + x²)¹¹ = ((1 + x)²)¹¹ = (1 + x)²², so the number of terms is 22 + 1 = 23. For the fourth expression (x + 2y - 3z)ⁿ = (x + (2y - 3z))ⁿ, we can write it as ⁿC₀xⁿ + ⁿC₁xⁿ⁻¹(2y - 3z)¹ + ⁿC₂xⁿ⁻²(2y - 3z)² + ... + ⁿCₙ(2y - 3z)ⁿ. The first term gives one term, the second term gives 2 terms, the third term gives 3 terms, and so on, with the last term giving (n + 1) terms. The total number of terms is 1 + 2 + 3 + ... + (n + 1) = \( \frac{(n+1)(n+2)}{2} \).
In simple words: For most binomial expressions (a + b)ⁿ, just add 1 to the power n to get the number of terms. When you have three variables like (x + 2y - 3z)ⁿ, the count is \( \frac{(n+1)(n+2)}{2} \) because each part breaks down into more pieces.

Exam Tip: Recognize whether the binomial is simple or complex. For expressions like (1 + 2x + x²), try to rewrite them as perfect squares or powers first - this often makes counting terms much easier. For three-variable expressions, use the triangular number formula.

Question 2. Expand the following:
(i) \( (3x - 2y)^4 \)
(ii) \( \left(x^2 + \frac{3}{x}\right)^4, x \neq 0 \)
Answer: (i) \( (3x - 2y)^4 = (3x + (-2y))^4 = {^4C_0}(3x)^4 + {^4C_1}(3x)^3(-2y) + {^4C_2}(3x)^2(-2y)^2 + {^4C_3}(3x)(-2y)^3 + {^4C_4}(-2y)^4 = 1 \cdot 81x^4 + 4 \cdot 27x^3 \cdot (-2y) + 6 \cdot 9x^2 \cdot 4y^2 + 4 \cdot 3x \cdot (-8y^3) + 1 \cdot 16y^4 = 81x^4 - 216x^3y + 216x^2y^2 - 96xy^3 + 16y^4 \).
(ii) \( \left(x^2 + \frac{3}{x}\right)^4 = {^4C_0}(x^2)^4 + {^4C_1}(x^2)^3\frac{3}{x} + {^4C_2}(x^2)^2\left(\frac{3}{x}\right)^2 + {^4C_3}x^2\left(\frac{3}{x}\right)^3 + {^4C_4}\left(\frac{3}{x}\right)^4 = x^8 + 4 \cdot x^6 \cdot \frac{3}{x} + 6 \cdot x^4 \cdot \frac{9}{x^2} + 4 \cdot x^2 \cdot \frac{27}{x^3} + \frac{81}{x^4} = x^8 + 12x^5 + 54x^2 + \frac{108}{x} + \frac{81}{x^4} \).
In simple words: Use the binomial formula by carefully applying each combination coefficient. Always change subtraction to adding a negative number to keep signs correct. Watch out with fractions - simplify powers carefully.

Exam Tip: Show all intermediate steps with coefficients written out. Examiners reward methodical substitution into the binomial formula far more than jumping to the final answer. Always verify your first and last terms - they should be the simplest ones.

Question 3. Expand the following:
(i) \( (2x^2 + 3y)^5 \)
(ii) \( \left(\frac{2x}{3} - \frac{3}{2x}\right)^4 \)
Answer: (i) \( (2x^2 + 3y)^5 = {^5C_0}(2x^2)^5 + {^5C_1}(2x^2)^4(3y) + {^5C_2}(2x^2)^3(3y)^2 + {^5C_3}(2x^2)^2(3y)^3 + {^5C_4}(2x^2)(3y)^4 + {^5C_5}(3y)^5 = 1 \cdot 2^5x^{10} + 5 \cdot 2^4 \cdot 3 \cdot x^8y + 10 \cdot 2^3 \cdot 3^2 \cdot x^6y^2 + 10 \cdot 2^2 \cdot 3^3 \cdot x^4y^3 + 5 \cdot 2 \cdot 3^4 \cdot x^2y^4 + 3^5y^5 = 32x^{10} + 240x^8y + 720x^6y^2 + 1080x^4y^3 + 810x^2y^4 + 243y^5 \).
(ii) \( \left(\frac{2x}{3} - \frac{3}{2x}\right)^4 = {^4C_0}\left(\frac{2x}{3}\right)^4 + {^4C_1}\left(\frac{2x}{3}\right)^3\left(-\frac{3}{2x}\right) + {^4C_2}\left(\frac{2x}{3}\right)^2\left(-\frac{3}{2x}\right)^2 + {^4C_3}\left(\frac{2x}{3}\right)\left(-\frac{3}{2x}\right)^3 + {^4C_4}\left(-\frac{3}{2x}\right)^4 = \frac{16x^8}{81} - 4 \cdot \frac{8x^6}{27} \cdot \frac{3}{2x} + 6 \cdot \frac{4x^4}{9} \cdot \frac{9}{4x^2} - 4 \cdot \frac{2x}{3} \cdot \frac{27}{8x^3} + \frac{81}{16x^4} = \frac{16x^8}{81} - \frac{16x^5}{9} + 6x^2 - \frac{9}{x} + \frac{81}{16x^4} \).
In simple words: Break down each term using the binomial formula. Calculate powers of fractions step by step - this keeps errors minimal. Combine powers of the same variable at the end.

Exam Tip: With fractional expressions, write out the general term first, then substitute your specific values. This catches sign errors early. Always simplify your final fractions.

Question 4. Expand the following:
(i) \( (3x^2 - 2ax + 3a^2)^3 \)
(ii) \( (1 - x + x^2)^4 \)
Answer: (i) We can rewrite (3x² - 2ax + 3a²)³ as (3(x² + a²) - 2ax)³ = {^3C_0}(3(x² + a²))³ - {^3C_1}(3(x² + a²))² \cdot 2ax + {^3C_2} \cdot 3(x² + a²) \cdot (2ax)² - {^3C_3}(2ax)³. Expanding this gives 27(x² + a²)³ - 3 \cdot 9(x² + a²)² \cdot 2ax + 3 \cdot 3(x² + a²) \cdot 4a²x² - 8a³x³. When we expand (x² + a²)³ = x⁶ + 3x⁴a² + 3x²a⁴ + a⁶ and (x² + a²)² = x⁴ + 2a²x² + a⁴, we get 27x⁶ + 81a²x⁴ + 81a⁴x² + 27a⁶ - 54ax⁵ - 108a³x³ - 54a⁵x + 36a²x⁴ + 36a⁴x² - 8a³x³ = 27x⁶ - 54ax⁵ + 117a²x⁴ - 116a³x³ + 117a⁴x² - 54a⁵x + 27a⁶.
(ii) We can write (1 - x + x²)⁴ as ((1 - x) + x²)⁴ = {^4C_0}(1 - x)⁴ + {^4C_1}(1 - x)³ \cdot x² + {^4C_2}(1 - x)² \cdot x⁴ + {^4C_3}(1 - x) \cdot x⁶ + {^4C_4}x⁸. Expanding (1 - x)⁴ = 1 - 4x + 6x² - 4x³ + x⁴, (1 - x)³ = 1 - 3x + 3x² - x³, and (1 - x)² = 1 - 2x + x², we get 1 · (1 - 4x + 6x² - 4x³ + x⁴) + 4(1 - 3x + 3x² - x³)x² + 6(1 - 2x + x²)x⁴ + 4(1 - x)x⁶ + x⁸ = 1 - 4x + 10x² - 16x³ + 19x⁴ - 16x⁵ + 10x⁶ - 4x⁷ + x⁸.
In simple words: When you see a trinomial, try to group it as a binomial plus another term. Then use the binomial expansion. Expand smaller pieces separately before combining them all together.

Exam Tip: Recognize that grouping a trinomial as (a + b + c) = (a + (b + c)) or ((a + b) + c) makes the binomial expansion work. Show your grouping step clearly so the examiner follows your logic.

Question 5. Using binomial theorem, find the values of:
(i) \( (99)^4 \)
(ii) \( (98)^5 \)
(iii) \( (1.02)^6 \) correct to 5 decimal places
Answer: (i) \( (99)^4 = (100 - 1)^4 = (10^2 - 1)^4 = {^4C_0}(10^2)^4 - {^4C_1}(10^2)^3 \cdot 1 + {^4C_2}(10^2)^2 \cdot 1^2 - {^4C_3}(10^2) \cdot 1^3 + {^4C_4} \cdot 1^4 = 1 \cdot 10^8 - 4 \cdot 10^6 + 6 \cdot 10^4 - 4 \cdot 10^2 + 1 = 100000000 - 4000000 + 60000 - 400 + 1 = 96059601 \).
(ii) \( (98)^5 = (100 - 2)^5 = (10^2 - 2)^5 = {^5C_0}(10^2)^5 - {^5C_1}(10^2)^4 \cdot 2 + {^5C_2}(10^2)^3 \cdot 2^2 - {^5C_3}(10^2)^2 \cdot 2^3 + {^5C_4}(10^2) \cdot 2^4 - {^5C_5} \cdot 2^5 = 1 \times 10^{10} - 5 \times 10^8 \times 2 + 10 \times 10^6 \times 4 - 10 \times 10^4 \times 8 + 5 \times 10^2 \times 16 - 1 \times 32 = 10000000000 - 1000000000 + 40000000 - 800000 + 8000 - 32 = 9039207968 \).
(iii) \( (1.02)^6 = (1 + 0.02)^6 = {^6C_0} + {^6C_1}(0.02) + {^6C_2}(0.02)^2 + {^6C_3}(0.02)^3 + {^6C_4}(0.02)^4 + {^6C_5}(0.02)^5 + {^6C_6}(0.02)^6 = 1 + 6(0.02) + 15(0.0004) + 20(0.000008) + 15(0.00000016) + 6(0.0000000032) + 1(0.000000000064) = 1 + 0.12 + 0.006 + 0.00016 + 0.0000024 + ... = 1.12616 \) correct to 5 decimal places.
In simple words: Express your number as something close to a power (like 100 - 1 or 1 + 0.02), then expand using the binomial formula. This turns hard arithmetic into simple multiplication and addition.

Exam Tip: Always write your number as (convenient base ± small adjustment) first. For decimal approximations, terms with higher powers become very tiny very fast - usually three or four terms are enough for good accuracy.

Question 6. Which number is larger: \( (1.2)^{4000} \) or 800?
Answer: \( (1.2)^{4000} = (1 + 0.2)^{4000} = {^{4000}C_0} + {^{4000}C_1}(0.2) + \text{other positive terms} = 1 + 4000(0.2) + \text{other positive terms} = 1 + 800 + \text{other positive terms} > 800 \). Therefore, \( (1.2)^{4000} > 800 \).
In simple words: When you expand (1.2)⁴⁰⁰⁰, the first two terms alone give you 1 + 800, and then you still have many more positive terms to add. So it must be much bigger than 800.

Exam Tip: For comparison questions, write the binomial expansion and look at just the first few terms. If those already exceed the target number, you are done - no need to calculate further.

Question 7. Find the value of \( (a^2 + \sqrt{a^2 - 1})^4 + (a^2 - \sqrt{a^2 - 1})^4 \).
Answer: Let \( b = \sqrt{a^2 - 1} \). Then \( (a^2 + \sqrt{a^2 - 1})^4 + (a^2 - \sqrt{a^2 - 1})^4 = (a^2 + b)^4 + (a^2 - b)^4 = ({^4C_0}(a^2)^4 + {^4C_1}(a^2)^3b + {^4C_2}(a^2)^2b^2 + {^4C_3}a^2b^3 + {^4C_4}b^4) + ({^4C_0}(a^2)^4 - {^4C_1}(a^2)^3b + {^4C_2}(a^2)^2b^2 - {^4C_3}a^2b^3 + {^4C_4}b^4) = 2({^4C_0}a^8 + {^4C_2}a^4b^2 + {^4C_4}b^4) = 2(a^8 + 6a^4(a^2 - 1) + (a^2 - 1)^2) = 2(a^8 + 6a^6 - 6a^4 + a^4 - 2a^2 + 1) = 2(a^8 + 6a^6 - 5a^4 - 2a^2 + 1) \).
In simple words: When you add expressions like (p + q)⁴ + (p - q)⁴, the odd-power terms cancel out. Only the even powers survive, making calculation much simpler.

Exam Tip: Recognize the (p + q)ⁿ + (p - q)ⁿ or (p + q)ⁿ - (p - q)ⁿ patterns - these always lead to nice cancellations. Use this to simplify before expanding.

Question 8. Expand \( (a + b)^6 - (a - b)^6 \). Hence find the value of \( (\sqrt{3} + \sqrt{2})^6 - (\sqrt{3} - \sqrt{2})^6 \).
Answer: \( (a + b)^6 - (a - b)^6 = ({^6C_0}a^6 + {^6C_1}a^5b + {^6C_2}a^4b^2 + {^6C_3}a^3b^3 + {^6C_4}a^2b^4 + {^6C_5}ab^5 + {^6C_6}b^6) - ({^6C_0}a^6 - {^6C_1}a^5b + {^6C_2}a^4b^2 - {^6C_3}a^3b^3 + {^6C_4}a^2b^4 - {^6C_5}ab^5 + {^6C_6}b^6) = 2({^6C_1}a^5b + {^6C_3}a^3b^3 + {^6C_5}ab^5) = 2(6a^5b + 20a^3b^3 + 6ab^5) = 4ab(3a^4 + 10a^2b^2 + 3b^4) \). Putting \( a = \sqrt{3} \) and \( b = \sqrt{2} \), we get \( (\sqrt{3} + \sqrt{2})^6 - (\sqrt{3} - \sqrt{2})^6 = 4\sqrt{3}\sqrt{2}[3(\sqrt{3})^4 + 10(\sqrt{3})^2(\sqrt{2})^2 + 3(\sqrt{2})^4] = 4\sqrt{6}(3 \times 9 + 10 \times 3 \times 2 + 3 \times 4) = 4\sqrt{6}(27 + 60 + 12) = 396\sqrt{6} \).
In simple words: When you subtract (a - b)⁶ from (a + b)⁶, all the even-power terms drop out. You are left with only odd-power terms, which is much easier to work with.

Exam Tip: Always use the (p + q)ⁿ - (p - q)ⁿ pattern to your advantage. Substitute the values only after you have simplified the general form - this prevents arithmetic mistakes.

Question 9. Using binomial theorem, evaluate \( (\sqrt{3} + 1)^5 - (\sqrt{3} - 1)^5 \). Hence show that the value of \( (\sqrt{3} + 1)^5 \) lies between 152 and 153.
Answer: \( (\sqrt{3} + 1)^5 - (\sqrt{3} - 1)^5 = ({^5C_0}(\sqrt{3})^5 + {^5C_1}(\sqrt{3})^4 + {^5C_2}(\sqrt{3})^3 + {^5C_3}(\sqrt{3})^2 + {^5C_4}(\sqrt{3}) + {^5C_5}) - ({^5C_0}(\sqrt{3})^5 - {^5C_1}(\sqrt{3})^4 + {^5C_2}(\sqrt{3})^3 - {^5C_3}(\sqrt{3})^2 + {^5C_4}(\sqrt{3}) - {^5C_5}) = 2({^5C_1}(\sqrt{3})^4 + {^5C_3}(\sqrt{3})^2 + {^5C_5}) = 2(5 \times 9 + 10 \times 3 + 1) = 2(45 + 30 + 1) = 152 \). So \( (\sqrt{3} + 1)^5 = 152 + (\sqrt{3} - 1)^5 \). Since \( \sqrt{3} = 1.732 \), we have \( 0 < \sqrt{3} - 1 < 1 \), which means \( 0 < (\sqrt{3} - 1)^5 < 1 \) (because if 0 < a < 1, then 0 < aⁿ < 1 for all n ∈ N). Therefore, from our equation, \( (\sqrt{3} + 1)^5 = 152 + (\text{a positive number less than 1}) \), which means \( (\sqrt{3} + 1)^5 \) lies between 152 and 153.
In simple words: Use the subtraction pattern to find a whole number. Then use that result plus a small positive remainder to show the answer is trapped between two integers.

Exam Tip: When asked to "show that" a value lies between two bounds, first show it equals an integer plus a remainder, then verify that the remainder is between 0 and 1.

Question 10. If P is the sum of odd-positioned terms and Q is the sum of even-positioned terms in the expansion of \( (x + a)^n \), prove that:
(i) \( P^2 - Q^2 = (x^2 - a^2)^n \)
(ii) \( 2(P^2 + Q^2) = (x + a)^{2n} + (x - a)^{2n} \)
(iii) \( 4PQ = (x + a)^{2n} - (x - a)^{2n} \)
Answer: We have \( (x + a)^n = {^nC_0}x^n + {^nC_1}x^{n-1}a + {^nC_2}x^{n-2}a^2 + {^nC_3}x^{n-3}a^3 + \ldots + {^nC_n}a^n = ({^nC_0}x^n + {^nC_2}x^{n-2}a^2 + \ldots) + ({^nC_1}x^{n-1}a + {^nC_3}x^{n-3}a^3 + \ldots) = P + Q \) ...(1). Similarly, \( (x - a)^n = {^nC_0}x^n - {^nC_1}x^{n-1}a + {^nC_2}x^{n-2}a^2 - {^nC_3}x^{n-3}a^3 + \ldots + {^nC_n}(-1)^na^n = ({^nC_0}x^n + {^nC_2}x^{n-2}a^2 + \ldots) - ({^nC_1}x^{n-1}a + {^nC_3}x^{n-3}a^3 + \ldots) = P - Q \) ...(2). (i) \( P^2 - Q^2 = (P + Q)(P - Q) = (x + a)^n(x - a)^n \) (using (1) and (2)) \( = ((x + a)(x - a))^n = (x^2 - a^2)^n \). (ii) \( 2(P^2 + Q^2) = (P + Q)^2 + (P - Q)^2 = ((x + a)^n)^2 + ((x - a)^n)^2 = (x + a)^{2n} + (x - a)^{2n} \) (using (1) and (2)). (iii) \( 4PQ = (P + Q)^2 - (P - Q)^2 = ((x + a)^n)^2 - ((x - a)^n)^2 = (x + a)^{2n} - (x - a)^{2n} \) (using (1) and (2)).
In simple words: Separating odd and even terms in an expansion, then using their sum and difference, gives you clever ways to build products and differences of complete expansions.

Exam Tip: This proof relies on knowing that (P + Q) and (P - Q) are the forward and backward expansions. Set them up early and use algebraic identities like a² - b² = (a + b)(a - b) to complete the proofs.

Question 11. Write the binomial expansion of \( (1 + x)^{n+1} \) when x = 8. Deduce that \( 9^{n+1} - 8n - 9 \) is divisible by 64 for all n ∈ N.
Answer: By the binomial theorem, \( (1 + x)^{n+1} = {^{n+1}C_0} + {^{n+1}C_1}x + {^{n+1}C_2}x^2 + {^{n+1}C_3}x^3 + \ldots + {^{n+1}C_{n+1}}x^{n+1} \). Putting x = 8, we get \( (1 + 8)^{n+1} = {^{n+1}C_0} + {^{n+1}C_1} \cdot 8 + {^{n+1}C_2} \cdot 8^2 + {^{n+1}C_3} \cdot 8^3 + \ldots + {^{n+1}C_{n+1}} \cdot 8^{n+1} \), which is the required binomial expansion. So \( 9^{n+1} = 1 + (n + 1) \cdot 8 + {^{n+1}C_2} \cdot 8^2 + {^{n+1}C_3} \cdot 8^3 + \ldots + {^{n+1}C_{n+1}} \cdot 8^{n+1} \) (since \( {^{n+1}C_0} = 1 \) and \( {^{n+1}C_1} = n + 1 \)). Therefore, \( 9^{n+1} - 8n - 9 = 1 + 8n + 8 + {^{n+1}C_2} \cdot 8^2 + {^{n+1}C_3} \cdot 8^3 + \ldots - 8n - 9 = 64({^{n+1}C_2} + {^{n+1}C_3} \cdot 8 + \ldots + {^{n+1}C_{n+1}} \cdot 8^{n-1}) = 64\lambda \), where λ is some integer. Therefore, \( 9^{n+1} - 8n - 9 \) is divisible by 64 for all n ∈ N.
In simple words: Expand (1 + 8)^(n+1) using the binomial formula with x = 8. The first two terms and the constant add to 9 + 8n. When you subtract 8n + 9, every remaining term has 64 as a factor.

Exam Tip: To prove divisibility using the binomial theorem, expand (1 + multiple)ⁿ, substitute, then show the result factors out the target divisor clearly.

Question 12. Using the Binomial theorem, prove that \( 3^{2n+2} - 8n - 9 \) is divisible by 64 for all natural numbers n.
Answer: We have \( 3^{2n+2} - 8n - 9 = (3^2)^{n+1} - 8n - 9 = 9^{n+1} - 8n - 9 = (1 + 8)^{n+1} - 8n - 9 = ({^{n+1}C_0} + {^{n+1}C_1} \cdot 8 + {^{n+1}C_2} \cdot 8^2 + {^{n+1}C_3} \cdot 8^3 + \ldots + {^{n+1}C_{n+1}} \cdot 8^{n+1}) - 8n - 9 = 1 + (n + 1) \cdot 8 + 8^2({^{n+1}C_2} + {^{n+1}C_3} \cdot 8 + \ldots + {^{n+1}C_{n+1}} \cdot 8^{n-1}) - 8n - 9 = 9 + 8n + 64({^{n+1}C_2} + {^{n+1}C_3} \cdot 8 + \ldots + {^{n+1}C_{n+1}} \cdot 8^{n-1}) - 8n - 9 = 64\lambda \), where λ is some integer. Therefore, \( 3^{2n+2} - 8n - 9 \) is divisible by 64 for all natural numbers n.
In simple words: Rewrite the base as 9 = (3²), then follow the same divisibility pattern as before. The 8² in the expansion naturally pulls out 64.

Exam Tip: Before expanding, simplify your bases. If you see 3^(2n+2), think 9^(n+1) immediately. This makes the binomial expansion much cleaner.

Question 13. Using binomial theorem, prove that \( 6^n - 5n \) always leaves the remainder 1 when divided by 25, for all n ∈ N.
Answer: We have \( 6^n - 5n = (1 + 5)^n - 5n = ({^nC_0} + {^nC_1} \times 5 + {^nC_2} \times 5^2 + {^nC_3} \times 5^3 + \ldots + {^nC_n} \times 5^n) - 5n = 1 + 5n + {^nC_2} \times 5^2 + {^nC_3} \times 5^3 + \ldots + {^nC_n} \times 5^n - 5n = 1 + 5^2({^nC_2} + {^nC_3} \times 5 + \ldots + {^nC_n} \times 5^{n-2}) = 1 + 25\lambda \), where \( \lambda = {^nC_2} + {^nC_3} \times 5 + \ldots + {^nC_n} \times 5^{n-2} \) is some integer. Therefore, \( 6^n - 5n = 25\lambda + 1 \), where λ is some integer. This means \( 6^n - 5n \) leaves the remainder 1 when divided by 25.
In simple words: Write 6 as 1 + 5, expand using the binomial formula, then notice that the 5n terms cancel. What remains is 1 plus a multiple of 25.

Exam Tip: For remainder problems, write your number as (1 + divisor factor)ⁿ. The expansion will naturally separate into "remainder" + "multiple of divisor".

Question 14. In the binomial expansion of \( (a + b)^n \), the (r + 1)th term is known as the general term.
Answer: In the binomial expansion of (a + b)ⁿ, we observe that the first term is \( {^nC_0}a^n \), the second term is \( {^nC_1}a^{n-1}b \), the third term is \( {^nC_2}a^{n-2}b^2 \), and so on. By looking at the pattern of successive terms, we discover that the (r + 1)th term is \( {^nC_r}a^{n-r}b^r \). This is called the general term and is denoted by \( T_{r+1} \). Hence, the general term is \( T_{r+1} = {^nC_r}a^{n-r}b^r \).
In simple words: The general term is a formula that lets you find any term in the expansion without writing out all the earlier ones. Use the position (r + 1) to calculate the coefficient and the powers directly.

Exam Tip: Remember that the general term \( T_{r+1} \) uses r to count from 0. So the 1st term has r = 0, the 2nd term has r = 1, and so on. This off-by-one relationship catches many students - be careful.

Question 15. State the particular cases of the general term formula.
Answer: (i) In the expansion of (a - b)ⁿ, the general term is \( T_{r+1} = (-1)^r {^nC_r}a^{n-r}b^r \). (ii) In the expansion of (1 + x)ⁿ, the general term is \( T_{r+1} = {^nC_r}x^r \). (iii) In the expansion of (1 - x)ⁿ, the general term is \( T_{r+1} = (-1)^r {^nC_r}x^r \).
In simple words: These three forms handle the most common types of binomial expansions. Notice how subtraction introduces the (-1)^r factor, and using 1 as one of the terms simplifies the powers.

Exam Tip: Memorize these three particular cases. Recognize which one matches your question and substitute directly - this saves time on calculations.

Question 16. Find the 7th term in the expansion of \( \left(2x^3 - \frac{3}{2x}\right)^{10} \).
Answer: In the expansion of (a + b)ⁿ, the general term is \( T_{r+1} = {^nC_r}a^{n-r}b^r \). In the expansion of \( \left(2x^3 - \frac{3}{2x}\right)^{10} \), we have \( T_7 = T_{6+1} = {^{10}C_6}(2x^3)^{10-6} \left(-\frac{3}{2x}\right)^6 = {^{10}C_4}(2x^3)^4 \left(\frac{3}{2x}\right)^6 = \frac{10 \times 9 \times 8 \times 7}{1 \times 2 \times 3 \times 4} \times 2^4 \times x^{12} \times \frac{3^6}{2^6 \times x^6} = 210 \times 16 \times x^{12} \times \frac{729}{64 \times x^6} = \frac{210 \times 16 \times 729}{64} \times x^6 = \frac{76545}{2}x^6 \).
In simple words: Use the general term formula. Substitute r = 6 (since we want the 7th term = T_{6+1}). Carefully calculate the combination, multiply out the powers, and simplify.

Exam Tip: Always double-check that r + 1 gives you the correct term number. Write out the combination carefully to avoid arithmetic mistakes with large factorials.

Question 17. Find the 13th term in the expansion of \( \left(9x - \frac{1}{3\sqrt{x}}\right)^{18}, x > 0 \).
Answer: In the expansion of \( \left(9x - \frac{1}{3\sqrt{x}}\right)^{18} \), we have \( T_{13} = T_{12+1} = {^{18}C_{12}}(9x)^{18-12} \left(-\frac{1}{3\sqrt{x}}\right)^{12} = {^{18}C_6}(9x)^6 \times \frac{1}{3^{12}x^6} \) (since \( {^nC_r} = {^nC_{n-r}} \)) \( = {^{18}C_6} \times 9^6 \times x^6 \times \frac{1}{3^{12} \times x^6} = \frac{18 \times 17 \times 16 \times 15 \times 14 \times 13}{1 \times 2 \times 3 \times 4 \times 5 \times 6} \times 3^{12} \times \frac{1}{3^{12}} = 18564 \).
In simple words: Set r = 12, then apply the general term formula. Use the property that C(n,r) = C(n, n-r) to simplify your combination. The x terms should cancel out if everything works correctly.

Exam Tip: For these more complex expressions, write the general term first, then substitute r. Verify that x powers do cancel - if they don't, you likely made a sign or exponent error.

Question 18. Find the rth term from the end in the expansion of \( (x + a)^n, n \in N \).
Answer: The rth term from the end in the expansion of (x + a)ⁿ equals the rth term from the beginning in the expansion of (a + x)ⁿ, which is \( T_r = T_{(r-1)+1} = {^nC_{r-1}}a^{n-(r-1)}x^{r-1} = {^nC_{r-1}}x^{r-1}a^{n-r+1} \).
In simple words: When you want a term from the end, swap the two parts of the binomial (x and a trade places), then find the same position from the start of that swapped expansion.

Exam Tip: You can also use: rth term from end = (n - r + 2)th term from beginning. Both approaches work - pick whichever makes your arithmetic simpler.

Question 19. Find the fourth term from the end in the expansion of \( \left(\frac{3}{x^2} - \frac{x^3}{3}\right)^9 \).
Answer: The fourth term from the end in the expansion of \( \left(\frac{3}{x^2} - \frac{x^3}{3}\right)^9 = \) the fourth term from the beginning in the expansion of \( \left(-\frac{x^3}{3} + \frac{3}{x^2}\right)^9 \). So \( T_4 = T_{3+1} = {^9C_3}\left(-\frac{x^3}{3}\right)^{9-3}\left(\frac{3}{x^2}\right)^3 = \frac{9 \times 8 \times 7}{1 \times 2 \times 3}\left(-\frac{x^3}{3}\right)^6\left(\frac{3}{x^2}\right)^3 = 84 \times \frac{x^{18}}{3^6} \times \frac{27}{x^6} = 84 \times \frac{27 \times x^{18}}{729 \times x^6} = 84 \times \frac{27}{729} \times x^{12} = 84 \times \frac{1}{27} \times x^{12} = \frac{28}{9}x^{12} \). Alternatively, 4th term from end = (9 - 4 + 2)th = 7th term from beginning. \( T_7 = T_{6+1} = {^9C_6}\left(\frac{3}{x^2}\right)^{9-6}\left(-\frac{x^3}{3}\right)^6 = {^9C_3}\left(\frac{3}{x^2}\right)^3\left(-\frac{x^3}{3}\right)^6 = 84 \times \frac{27}{x^6} \times \frac{x^{18}}{729} = \frac{28}{9}x^{12} \).
In simple words: Swap the order and find from the start. Or use the formula: position from end = (n - position + 2) from start. Both give the same answer.

Exam Tip: When finding terms from the end, always verify your answer using both methods - they serve as a check on each other.

Question 20. Find x if the 17th and 18th terms of the expansion \( (2 + x)^{50} \) are equal.
Answer: In the expansion of (2 + x)⁵⁰, we have \( T_{17} = T_{16+1} = {^{50}C_{16}}2^{50-16}x^{16} \) and \( T_{18} = T_{17+1} = {^{50}C_{17}}2^{50-17}x^{17} \). Given that \( T_{17} = T_{18} \), we have \( {^{50}C_{16}}2^{34}x^{16} = {^{50}C_{17}}2^{33}x^{17} \). Dividing both sides by \( 2^{33}x^{16} \), we get \( {^{50}C_{16}} \times 2 = {^{50}C_{17}} \times x \). So \( \frac{{^{50}C_{16}}}{{^{50}C_{17}}} \times 2 = x \). Now \( \frac{{^{50}C_{16}}}{{^{50}C_{17}}} = \frac{50!/(16! \times 34!)}{50!/(17! \times 33!)} = \frac{50! \times 17! \times 33!}{16! \times 34! \times 50!} = \frac{17}{34} = \frac{1}{2} \). Therefore, \( x = \frac{1}{2} \times 2 = 1 \).
In simple words: Set up the two general terms using r = 16 and r = 17. Since they are equal, write an equation and simplify by canceling. The combination ratio simplifies to a simple fraction.

Exam Tip: When consecutive terms are equal, use the general term formula for both, then divide them to create a ratio. The x's and most powers cancel, leaving a clean equation.

Question 21. Find the middle term in \( \left(\frac{2x^2}{3} - \frac{3}{2x}\right)^{12} \).
Answer: The total number of terms in the given expansion is 12 + 1 = 13 (odd). Therefore, there is only one middle term given by \( T_{12/2 + 1} \) i.e., \( T_7 \). \( T_7 = T_{6+1} = {^{12}C_6}\left(\frac{2x^2}{3}\right)^{12-6}\left(-\frac{3}{2x}\right)^6 = {^{12}C_6}\left(\frac{2x^2}{3}\right)^6\left(\frac{3}{2x}\right)^6 = \frac{18 \times 17 \times 16 \times 15 \times 14 \times 13}{1 \times 2 \times 3 \times 4 \times 5 \times 6} \times 2^6 \times x^{12} \times \frac{3^6}{3^6 \times 2^6 \times x^6} = 18564 \times x^6 \times \frac{1}{x^6} = 18564 \).
In simple words: Count the terms: 12 + 1 = 13, which is odd. The middle one is the 7th. Use the general term formula with r = 6 to find T₇.

Exam Tip: For odd total terms, the middle is at position (n + 2)/2. For even total terms, there are two middle positions. Always count the total terms first.