RS Aggarwal Solutions for Class 11 Chapter 13 Some Special Series

Question 1. Find the sum of the series whose nth term is given by: (3n² + 2n)
Answer: The nth term of the series is a_n = 3n² + 2n. To determine the sum S_n, we expand the summation as follows:

S_n = Σ(3n² + 2n) from n=1 to n

= 3Σ(n²) + 2Σ(n)

Applying the standard formulas:

Σ(n) = n(n+1)/2

Σ(n²) = n(n+1)(2n+1)/6

We get:

S_n = 3 · [n(n+1)(2n+1)/6] + 2 · [n(n+1)/2]

= [n(n+1)(2n+1)]/2 + n(n+1)

= n(n+1)[(2n+1)/2 + 1]

= n(n+1)[(2n+3)/2]

= [n(n+1)(2n+3)]/2

Exam Tip: Always break down the general term into simpler components and apply the standard summation formulas methodically. Verify your final expression by substituting small values of n.

Question 2. Find the sum of the series whose nth term is given by: n(n+1)(n+4)
Answer: Given that a_n = n(n+1)(n+4), we need to find S_n. First, expand the general term:

a_n = n(n+1)(n+4) = n(n² + 5n + 4) = n³ + 5n² + 4n

Thus:

S_n = Σ(n³ + 5n² + 4n) from n=1 to n

= Σ(n³) + 5Σ(n²) + 4Σ(n)

Using the standard identities:

Σ(n) = n(n+1)/2

Σ(n²) = n(n+1)(2n+1)/6

Σ(n³) = [n(n+1)/2]²

We get:

S_n = [n(n+1)/2]² + 5 · [n(n+1)(2n+1)/6] + 4 · [n(n+1)/2]

After simplification, the sum of the series is:

S_n = [n(n+1)(n² + 6n + 8)]/4 = [n(n+1)(n+2)(n+4)]/4

Exam Tip: When the general term is a product of linear factors, expand and separate into sum of individual power terms. This allows you to apply each standard formula independently.

Question 3. Find the sum of the series whose nth term is given by: (4n³ + 6n² + 2n)
Answer: Given a_n = 4n³ + 6n² + 2n, we need to find S_n:

S_n = Σ(4n³ + 6n² + 2n) from n=1 to n

= 4Σ(n³) + 6Σ(n²) + 2Σ(n)

Applying standard formulas:

S_n = 4 · [n(n+1)/2]² + 6 · [n(n+1)(2n+1)/6] + 2 · [n(n+1)/2]

= (n(n+1))²/4 + 5 · [n(n+1)(2n+1)/6] + 2n(n+1)

Factor out n(n+1):

S_n = n(n+1) · {[n(n+1)]/4 + 5[(2n+1)/6] + 2}

= [n(n+1)/24] · (6n(n+1) + 20(2n+1) + 48)

= [n(n+1)/12] · (6n² + 46n + 68)

= [n(n+1)/12] · (6(n² + 23n/3 + 34/3))

The sum of the series is: S_n = [n(n+1)(6n² + 46n + 68)]/24 = [n(n+1)(3n² + 23n + 34)]/12

Exam Tip: Factor out common terms after substitution to simplify the expression. Always double-check by expanding back to verify correctness.

Question 4. Find the sum of the series whose nth term is given by: (3n² - 3n + 2)
Answer: Given a_n = 3n² - 3n + 2, we determine S_n:

S_n = Σ(3n² - 3n + 2) from n=1 to n

= 3Σ(n²) - 3Σ(n) + Σ(2)

Applying formulas:

S_n = 3 · [n(n+1)(2n+1)/6] - 3 · [n(n+1)/2] + 2n

= [n(n+1)(2n+1)]/2 - 3n(n+1)/2 + 2n

Combine over a common denominator:

= [n(n+1)(2n+1) - 3n(n+1) + 4n]/2

= [n((n+1)(2n+1) - 3(n+1) + 4)]/2

= [n(n+1)((2n+1) - 3) + 4n]/2

= [n(n+1)(2n - 2) + 4n]/2

= n[(n+1) · 2(n-1) + 4]/2

After simplification: S_n = n(n+1)²(n+2)

Exam Tip: When combining fractions, find the common denominator first, then factor carefully. Factoring out common terms often reveals the final simplified form quickly.

Question 5. Find the sum of the series whose nth term is given by: (2n² - 3n + 5)
Answer: Given a_n = 2n² - 3n + 5, we compute S_n:

S_n = Σ(2n² - 3n + 5) from n=1 to n

= 2Σ(n²) - 3Σ(n) + 5Σ(1)

= 2 · [n(n+1)(2n+1)/6] - 3 · [n(n+1)/2] + 5n

= [n(n+1)(2n+1)]/3 - 3n(n+1)/2 + 5n

Converting to a common denominator (6):

= [2n(n+1)(2n+1) - 9n(n+1) + 30n]/6

= [n(2(n+1)(2n+1) - 9(n+1) + 30)]/6

= [n(4n² + 6n + 2 - 9n - 9 + 30)]/6

= [n(4n² - 3n + 23)]/6

The sum of the series is: S_n = [n(4n² - 3n + 23)]/6

Exam Tip: Work systematically through each summation term, apply the correct formula, then combine using a common denominator. Simplify step-by-step to avoid algebraic errors.

Question 6. Find the sum of the series whose nth term is given by: (n³ - 3ⁿ)
Answer: Given a_n = n³ - 3ⁿ, we need to find S_n:

S_n = Σ(n³ - 3ⁿ) from n=1 to n

= Σ(n³) - Σ(3ⁿ)

For the first part:

Σ(n³) = [n(n+1)/2]²

For the second part, Σ(3ⁿ) from n=1 to n is a geometric series with first term a = 3 and common ratio r = 3:

Σ(3ⁿ) = 3 + 3² + 3³ + ... + 3ⁿ = [3(3ⁿ - 1)]/(3 - 1) = [3(3ⁿ - 1)]/2

Therefore:

The sum of the series is: S_n = [n(n+1)/2]² - [3(3ⁿ - 1)]/2 = [n²(n+1)²]/4 - [3(3ⁿ - 1)]/2

Exam Tip: Recognize when a portion of the series forms a geometric progression. Use the geometric series formula separately, then combine with polynomial sum results.

Question 7. Find the sum of the series: (2² + 4² + 6² + 8² + ... to n terms)
Answer: The given series is 2² + 4² + 6² + 8² + ... to n terms. This can be rewritten as [(2·1)², (2·2)², (2·3)², ..., (2·n)²]. Therefore, the nth term is:

a_n = (2n)² = 4n²

The sum becomes:

S_n = Σ(4n²) from n=1 to n = 4Σ(n²)

Using the standard formula Σ(n²) = n(n+1)(2n+1)/6:

S_n = 4 · [n(n+1)(2n+1)/6] = [2n(n+1)(2n+1)]/3

The sum of the series is: S_n = [2n(n+1)(2n+1)]/3

Exam Tip: When dealing with series of even numbers, odd numbers, or multiples, factor out the multiplier and express in terms of natural numbers. This simplifies application of standard formulas.

Question 8. Find the sum of the series: (2³ + 4³ + 6³ + 8³ + ... to n terms)
Answer: The given series is 2³ + 4³ + 6³ + 8³ + ... to n terms. This can be rewritten as [(2·1)³, (2·2)³, (2·3)³, ..., (2·n)³]. Therefore, the nth term is:

a_n = (2n)³ = 8n³

The sum becomes:

S_n = Σ(8n³) from n=1 to n = 8Σ(n³)

Using the standard formula Σ(n³) = [n(n+1)/2]²:

S_n = 8 · [n(n+1)/2]² = 8 · [n²(n+1)²]/4 = 2n²(n+1)²

The sum of the series is: S_n = 2n²(n+1)²

Exam Tip: For series of powers of multiples, extract the constant multiplier, raise it to the appropriate power, then apply standard summation formulas. Verify results by computing the first few terms.

Question 9. Find the sum of the series: (5² + 6² + 7² + … + 20²)
Answer: To solve this question, we must first determine the nth term of the series so that we can apply the summation of the series using standard identities to obtain the required sum. The given series is 5², 6², 7² … 20². The series can be expressed as [(1 + 4)², (2 + 4)², (3 + 4)² … (16 + 4)²]. Thus, the nth term of the series is:
\( a_n = (n + 4)^2 \)
With n = 16,
\( a_n = n^2 + 8n + 16 \) Now, to find the sum of this series, \( S_n \):
\( S_n = \sum_{n=1}^{n} a_n \)
\( S_n = \sum_{n=1}^{n} (n^2 + 8n + 16) \)
\( = \sum_{n=1}^{n} (n^2) + \sum_{n=1}^{n} (8n) + \sum_{n=1}^{n} (16) \) From the standard formulas:
\( \sum_{n=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \)
\( \sum_{n=1}^{n} k = \frac{n(n+1)}{2} \)
\( \sum_{n=1}^{N} k = Nk \) Thus, for the given series:
\( S_n = \frac{n(n+1)(2n+1)}{6} + 8 \left( \frac{n(n+1)}{2} \right) + 16n \) Substituting n = 16:
\( S_n = \frac{(16)(17)(33)}{6} + 8 \left( \frac{(16)(17)}{2} \right) + 16 \times 16 \)
\( S_n = 1496 + 1088 + 256 = 2840 \) The sum of the series is \( S_n = 2840 \).
In simple words: To find this sum, express each term using a pattern, break down the series into parts using standard sum formulas, and then substitute the value n = 16 to get the final answer of 2840.

Exam Tip: Always identify the general term pattern first and decompose complicated series into simpler summation formulas - this approach works consistently for polynomial series.

Question 10. Find the sum of the series: (1 × 2) + (2 × 3) + (3 × 4) + (4 × 5) + … to n terms
Answer: To solve this question, we must first determine the nth term of the series so that we can apply the summation of the series using standard identities to obtain the required sum. The given series is (1 × 2) + (2 × 3) + (3 × 4) + (4 × 5) + … to n terms. The series can be expressed as [1 × (1 + 1), 2 × (2 + 1), 3 × (3 + 1), … n × (n + 1)]. Thus, the nth term of the series is:
\( a_n = n(n + 1) \)
\( a_n = n^2 + n \) Now, to find the sum of this series, \( S_n \):
\( S_n = \sum_{n=1}^{n} a_n \)
\( S_n = \sum_{n=1}^{n} (n^2 + n) \)
\( = \sum_{n=1}^{n} (n^2) + \sum_{n=1}^{n} (n) \) From the standard formulas:
\( S_n = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \)
\( S_n = \left( \frac{n(n+1)}{2} \right) \left( \frac{2n+4}{3} \right) \)
\( S_n = \left( \frac{n(n+1)}{2} \right) \left( \frac{2n+4}{3} \right) \) The sum of the series is \( S_n = \left( \frac{n(n+1)}{2} \right) \left( \frac{2n+4}{3} \right) \).
In simple words: Break down the series term into n squared plus n, apply the sum formulas for each part separately, and combine them together to get the final expression.

Exam Tip: Recognize that many series can be split into component polynomial parts - once split, use standard sum identities for each component rather than searching for a single complex formula.

Question 11. Find the sum of the series: (3 × 8) + (6 × 11) + (9 × 14) + … to n terms
Answer: To solve this question, we must first determine the nth term of the series so that we can apply the summation of the series using standard identities to obtain the required sum. The given series is (3 × 8) + (6 × 11) + (9 × 14) + … to n terms. The series can be expressed as [(3 × 1) × (3 × 1 + 5), (3 × 2) × (3 × 2 + 5), … (3n) × (3n + 5)]. Thus, the nth term of the series is:
\( a_n = 3n(3n + 5) \)
\( a_n = 9n^2 + 15n \) Now, to find the sum of this series, \( S_n \):
\( S_n = \sum_{n=1}^{n} (9n^2 + 15n) \) From the standard formulas:
\( S_n = 9 \sum_{n=1}^{n} (n^2) + 15 \sum_{n=1}^{n} (n) \)
\( S_n = 9 \left( \frac{n(n+1)(2n+1)}{6} \right) + 15 \left( \frac{n(n+1)}{2} \right) \)
\( S_n = \left( \frac{n(n+1)}{2} \right) \left( \frac{2n+4}{3} \right) \) The sum of the series is \( S_n = \left( \frac{n(n+1)(n+2)}{3} \right) \).
In simple words: Identify the general term as a product of two linear expressions in n, expand it into polynomial form, separate it into standard sum formulas, and then combine the results to get your answer.

Exam Tip: When multiplying two linear expressions like (3n) and (3n + 5), always expand completely before breaking into summation parts - this prevents errors in later calculations.

Question 12. Find the sum of the series: (1 × 2²) + (2 × 3²) + (3 × 4²) + … to n terms
Answer: To solve this question, we must first determine the nth term of the series so that we can apply the summation of the series using standard identities to obtain the required sum. The given series is (1 × 2²) + (2 × 3²) + (3 × 4²) + … to n terms. The series can be expressed as [1 × (1 + 1)², 2 × (2 + 1)², … n × (n + 1)²]. Thus, the nth term of the series is:
\( a_n = n(n + 1)^2 \)
\( a_n = n^3 + 2n^2 + n \) Now, to find the sum of this series, \( S_n \):
\( S_n = \sum_{n=1}^{n} a_n \)
\( S_n = \sum_{n=1}^{n} (n^3 + 2n^2 + n) \) From the standard formulas:
\( S_n = \sum_{n=1}^{n} (n^3) + 2 \sum_{n=1}^{n} (n^2) + \sum_{n=1}^{n} (n) \)
\( S_n = \left( \frac{n(n+1)}{2} \right)^2 + 2 \left( \frac{n(n+1)(2n+1)}{6} \right) + \left( \frac{n(n+1)}{2} \right) \)
\( = \left( \frac{n(n+1)}{2} \right) \left[ \frac{n(n+1)}{2} + \frac{2(2n+1)}{3} + 1 \right] \)
\( = \left( \frac{n(n+1)}{2} \right) \left[ \frac{3n^2 + 11n + 10}{6} \right] \) The sum of the series is \( S_n = \left( \frac{n(n+1)}{2} \right) \left[ \frac{3n^2 + 11n + 10}{6} \right] \).
In simple words: Expand n times (n + 1) squared to get three separate polynomial terms, apply the sum formula for each part individually, then combine them by factoring out common expressions to get the final simplified form.

Exam Tip: After expanding and summing each term, always look for common factors you can pull out - this simplification is often what separates a correct answer from an overly complicated one.

Question 13. Find the sum of the series: (1 × 2²) + (3 × 3²) + (5 × 4²) + … to n terms
Answer: To solve this question, we must first determine the nth term of the series so that we can apply the summation of the series using standard identities to obtain the required sum. The given series is (1 × 2²) + (3 × 3²) + (5 × 4²) + … to n terms. The series can be expressed as [1 × (1 + 1)², 2 × (2 + 1)², … (2n - 1) × (n + 1)²]. Thus, the nth term of the series is:
\( a_n = (2n - 1)(n + 1)^2 \)
\( = (2n - 1)(n^2 + 2n + 1) \)
\( = 2n^3 + 3n^2 - 1 \) Now, to find the sum of this series, \( S_n \):
\( S_n = \sum_{n=1}^{n} a_n \)
\( S_n = \sum_{n=1}^{n} (2n^3 + 3n^2 - 1) \) From the standard formulas:
\( S_n = 2 \sum_{n=1}^{n} (n^3) + 3 \sum_{n=1}^{n} (n^2) - \sum_{n=1}^{n} (1) \)
\( S_n = 2 \left( \frac{n(n+1)}{2} \right)^2 + 3 \left( \frac{n(n+1)(2n+1)}{6} \right) - n \)
\( = \left( \frac{n(n+1)}{2} \right) [n(n+1) + (2n+1)] - n \)
\( = \left( \frac{n(n+1)}{2} \right) [n^2 + 3n + 1] - n \)
\( = \left( \frac{n}{2} \right) [(n+1)(n^2 + 3n + 1) - 2] \)
\( = \left( \frac{n}{2} \right) [n^3 + 4n^2 + 4n - 1] \) The sum of the series is \( S_n = \left( \frac{n}{2} \right) [n^3 + 4n^2 + 4n - 1] \).
In simple words: Expand the product (2n - 1) times (n + 1) squared carefully to obtain three polynomial pieces, separate the sum into three parts using standard identities, and then gather the resulting expressions by taking out common factors.

Exam Tip: When the first multiplier alternates (1, 3, 5, …), write it explicitly as 2n - 1 from the start - this prevents sign errors and makes the expansion clearer.

Question 14. Find the sum of the series: (3 × 1²) + (5 × 2²) + (7 × 3²) + … to n terms
Answer: To solve this question, we must first determine the nth term of the series so that we can apply the summation of the series using standard identities to obtain the required sum. The given series is (3 × 1²) + (5 × 2²) + (7 × 3²) + … to n terms. The series can be expressed as [(3 × 1²), (5 × 2²), … ((2n + 1) × n²)]. Thus, the nth term of the series is:
\( a_n = (2n + 1)n^2 \)
\( a_n = 2n^3 + n^2 \) Now, to find the sum of this series, \( S_n \):
\( S_n = \sum_{n=1}^{n} a_n \)
\( S_n = \sum_{n=1}^{n} (2n^3 + n^2) \) From the standard formulas:
\( S_n = 2 \sum_{n=1}^{n} (n^3) + \sum_{n=1}^{n} (n^2) \)
\( S_n = 2 \left( \frac{n(n+1)}{2} \right)^2 + \frac{n(n+1)(2n+1)}{6} \)
\( = \left( \frac{n(n+1)}{2} \right) \left[ n(n+1) + \frac{2n+1}{3} \right] \)
\( = \left( \frac{n(n+1)}{2} \right) \left[ \frac{3n^2 + 5n + 1}{3} \right] \)
\( = \left( \frac{n(n+1)}{6} \right) (3n^2 + 5n + 1) \) The sum of the series is \( S_n = \left( \frac{n(n+1)}{6} \right) (3n^2 + 5n + 1) \).
In simple words: Break the general term into 2n cubed plus n squared, use the cube and square sum formulas separately, then factor out the common part n times (n + 1) divided by 2 to reach the final simplified form.

Exam Tip: Notice that the coefficients (3, 5, 7, …) form an arithmetic sequence - identifying this pattern early helps you write the general term correctly as 2n + 1.

Question 15. Find the sum of the series: (1 × 2 × 3) + (2 × 3 × 4) + (3 × 4 × 5) + … to n terms
Answer: To solve this question, we must first determine the nth term of the series so that we can apply the summation of the series using standard identities to obtain the required sum. The given series is (1 × 2 × 3) + (2 × 3 × 4) + (3 × 4 × 5) + … to n terms. The series can be expressed as [1 × (1 + 1) × (1 + 2), 2 × (2 + 1) × (2 + 2), … n × (n + 1) × (n + 2)]. Thus, the nth term of the series is:
\( a_n = n(n + 1)(n + 2) \)
\( a_n = n^3 + 3n^2 + 2n \) Now, to find the sum of this series, \( S_n \):
\( S_n = \sum_{n=1}^{n} a_n \)
\( S_n = \sum_{n=1}^{n} (n^3 + 3n^2 + 2n) \) From the standard formulas:
\( S_n = \sum_{n=1}^{n} (n^3) + 3 \sum_{n=1}^{n} (n^2) + 2 \sum_{n=1}^{n} (n) \)
\( S_n = \left( \frac{n(n+1)}{2} \right)^2 + 3 \left( \frac{n(n+1)(2n+1)}{6} \right) + 2 \left( \frac{n(n+1)}{2} \right) \)
\( = \left( \frac{n(n+1)}{2} \right) \left[ \frac{n(n+1)}{2} + (2n+1) + 2 \right] \)
\( = \left( \frac{n(n+1)}{2} \right) \left[ \frac{n^2 + 5n + 6}{2} \right] \)
\( = \left( \frac{n(n+1)}{2} \right) \left[ \frac{(n+3)(n+2)}{2} \right] \) The sum of the series is \( S_n = \left( \frac{n(n+1)(n+2)(n+3)}{4} \right) \).
In simple words: Expand the product of three consecutive natural numbers multiplied by n into separate cubic, quadratic, and linear terms, apply the standard sum formulas to each part, and then factor the final expression to reveal the elegant pattern of four consecutive factors divided by 4.

Exam Tip: Products of consecutive integers often yield formulas involving even more consecutive integers - factor aggressively at the end to reveal this beautiful pattern, which is a hallmark of correct solutions for this type of series.

Question 16. Find the sum of the series: (1 × 2 × 4) + (2 × 3 × 7) + (3 × 4 × 10) + … to n terms
Answer: To determine the sum, we first identify the general term. Observing the pattern, each group can be expressed as \( n(n+1)(3n+1) \), which expands to \( 3n^3 + 4n^2 + n \).

The sum of the series is:
\[ S_n = \sum_{n=1}^{n} (3n^3 + 4n^2 + n) = 3\sum n^3 + 4\sum n^2 + \sum n \]

Using standard formulas:
\[ S_n = 3\left(\frac{n(n+1)}{2}\right)^2 + 4\left(\frac{n(n+1)(2n+1)}{6}\right) + \frac{n(n+1)}{2} \]

Simplifying:
\[ S_n = \frac{n(n+1)}{2}\left[\frac{3n(n+1)}{2} + \frac{4(2n+1)}{3} + 1\right] = \frac{n(n+1)}{2}\left[\frac{9n^2 + 25n + 14}{6}\right] \]

\[ S_n = \frac{n(n+1)(9n^2 + 25n + 14)}{12} \]
In simple words: To add up all the terms, first write down what the nth term looks like. Then use known formulas for sums of powers to combine everything into one final expression.

Exam Tip: Always expand the general term and separate it into polynomial components - this lets you apply standard summation identities directly.

Question 17. Find the sum of the series: \( \frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \ldots \) to n terms
Answer: The general term of this series is \( a_n = \frac{1}{n(n+1)} \). Using partial fractions, we decompose this as:
\[ a_n = \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} \]

When we add all terms from the first to the nth:
\[ S_n = \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \cdots + \left(\frac{1}{n} - \frac{1}{n+1}\right) \]

Most terms cancel out (telescoping series), leaving:
\[ S_n = 1 - \frac{1}{n+1} = \frac{n}{n+1} \]
In simple words: Break each fraction into two parts. When you add them all up, most pieces cancel out, and you are left with just the first and last pieces.

Exam Tip: Partial fractions with consecutive denominators almost always lead to telescoping - watch for terms canceling from the middle.

Question 18. Find the sum of the series: \( \frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \cdots + \frac{1}{(2n-1)(2n+1)} \)
Answer: The general term is \( a_n = \frac{1}{(2n-1)(2n+1)} \). Using partial fractions:
\[ a_n = \frac{1}{(2n-1)(2n+1)} = \frac{1}{2}\left(\frac{1}{2n-1} - \frac{1}{2n+1}\right) \]

Writing out all terms:
\[ S_n = \frac{1}{2}\left[\left(\frac{1}{1} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \left(\frac{1}{5} - \frac{1}{7}\right) + \cdots + \left(\frac{1}{2n-1} - \frac{1}{2n+1}\right)\right] \]

Through telescoping, consecutive fractions eliminate each other:
\[ S_n = \frac{1}{2}\left(1 - \frac{1}{2n+1}\right) = \frac{1}{2} \cdot \frac{2n}{2n+1} = \frac{n}{2n+1} \]
In simple words: Split each term using partial fractions. When added together, inner terms vanish, and you keep only the very first and very last pieces.

Exam Tip: In telescoping series with odd denominators, check the coefficient in front - it often simplifies the final answer.

Question 19. Find the sum of the series: (1 × 2²) + (2 × 3²) + (3 × 4²) + … to n terms
Answer: The general term of this series is \( a_n = n(n+1)^2 = n^3 + 2n^2 + n \).

The sum becomes:
\[ S_n = \sum_{n=1}^{n} (n^3 + 2n^2 + n) = \sum n^3 + 2\sum n^2 + \sum n \]

Applying standard summation formulas:
\[ S_n = \left(\frac{n(n+1)}{2}\right)^2 + 2\left(\frac{n(n+1)(2n+1)}{6}\right) + \frac{n(n+1)}{2} \]

Factoring out \( \frac{n}{2} \):
\[ S_n = \frac{n}{2}\left[\frac{n(n+1)}{2} + \frac{n(n+1)(2n+1)}{3} + (n+1)\right] \]

Simplifying the bracket:
\[ S_n = \frac{n}{2}\left[\frac{n+1}{2}(3n + 4) + (n+1)\right] = \frac{n}{2(2n+1)} \]

\[ S_n = \frac{n}{2n+1} \]
In simple words: Write the general term as a sum of powers. Apply the formulas you know for summing first, second, and third powers. Combine and factor to get a clean final answer.

Exam Tip: Always expand polynomial terms fully before splitting into separate summations - this makes applying standard formulas straightforward.

Question 20. Find the sum of the series: \( \frac{1^3}{1} + \frac{1^3 + 2^3}{1 + 3} + \frac{1^3 + 2^3 + 3^3}{1 + 3 + 5} + \ldots \) to n terms
Answer: For the general term, note that the denominator is the sum of the first n odd numbers. Using the formula for sum of cubes, the numerator is \( \left[\frac{n(n+1)}{2}\right]^2 \), and the denominator equals \( n^2 \).

Therefore, the general term is:
\[ a_n = \frac{\left[\frac{n(n+1)}{2}\right]^2}{n^2} = \frac{(n+1)^2}{4} = \frac{n^2 + 2n + 1}{4} \]

The series sum becomes:
\[ S_n = \sum_{n=1}^{n} \frac{n^2 + 2n + 1}{4} = \frac{1}{4}\left[\sum n^2 + 2\sum n + \sum 1\right] \]

Applying summation identities:
\[ S_n = \frac{1}{4}\left[\frac{n(n+1)(2n+1)}{6} + 2\frac{n(n+1)}{2} + n\right] \]

\[ S_n = \frac{n}{4}\left[\frac{(n+1)(2n+1)}{6} + (n+1) + 1\right] = \frac{n(3n^2 + 11n + 10)}{12} \]
In simple words: Recognize that the numerator uses a formula for summing cubes and the denominator is the sum of odd numbers (which is always a perfect square). This makes the general term simpler than it looks.

Exam Tip: Look for hidden formulas in series numerators and denominators - sum of cubes, sum of odd numbers, and sum of an AP are common patterns.

Question 21. Find the sum of the series: 3 + 15 + 35 + 63 + … to n terms
Answer: The terms can be rewritten as \( 2^2 - 1, 4^2 - 1, 6^2 - 1, \ldots, (2n)^2 - 1 \). The general term is:
\[ a_n = (2n)^2 - 1 = 4n^2 - 1 \]

The series sum is:
\[ S_n = \sum_{n=1}^{n} (4n^2 - 1) = 4\sum n^2 - \sum 1 \]

Using standard formulas:
\[ S_n = 4 \cdot \frac{n(n+1)(2n+1)}{6} - n = \frac{2n(n+1)(2n+1)}{3} - n \]

\[ S_n = \frac{n}{3}[2(n+1)(2n+1) - 3] = \frac{n}{3}[4n^2 + 6n + 2 - 3] \]

\[ S_n = \frac{n(4n^2 + 6n - 1)}{3} \]
In simple words: Write each term as a perfect square minus one. Split the sum into two parts - one for the squares and one for the constants. Use the sum-of-squares formula to finish.

Exam Tip: When terms look like differences of powers, always express them in factored form first - it often reveals cancellations or standard patterns.

Question 22. Find the sum of the series: 1 + 5 + 12 + 22 + 35 + … to n terms
Answer: Using the method of differences, we examine the sequence of differences between consecutive terms: 4, 7, 10, 13, …. These form an arithmetic sequence with first term 1 and common difference 3.

The general term of the original series is the sum of the arithmetic sequence 1, 4, 7, 10, … up to n terms:
\[ a_n = 1 + 4 + 7 + 10 + \cdots + [1 + 3(n-1)] \]

This equals:
\[ a_n = \frac{n}{2}[2(1) + (n-1)(3)] = \frac{n}{2}[2 + 3n - 3] = \frac{n(3n - 1)}{2} \]

Now the series sum is:
\[ S_n = \sum_{n=1}^{n} \frac{n(3n - 1)}{2} = \frac{1}{2}\left[3\sum n^2 - \sum n\right] \]

\[ S_n = \frac{1}{2}\left[3 \cdot \frac{n(n+1)(2n+1)}{6} - \frac{n(n+1)}{2}\right] \]

\[ S_n = \frac{n(n+1)}{4}\left[(2n+1) - 1\right] = \frac{n^2(n+1)}{2} \]
In simple words: Look at how each term changes from the one before it. If those changes follow a pattern (like arithmetic), you can work backward to find the original terms, then sum everything.

Exam Tip: The method of differences works best when first differences form an AP or GP - always compute them first as a diagnostic step.

Question 23. Find the sum of the series: 5 + 7 + 13 + 31 + 85 + … to n terms
Answer: Using the method of differences, compute the sequence of differences: 2, 6, 18, 54, …. These form a geometric sequence with first term 2 and common ratio 3.

Using the difference method, the general term of the original series is:
\[ a_n = 5 + 2 + 6 + 18 + \cdots + [2 \cdot 3^{n-2}] \]

The sum of the geometric part (for n ≥ 2) is:
\[ 2 + 6 + 18 + \cdots + [2 \cdot 3^{n-2}] = 2(1 + 3 + 3^2 + \cdots + 3^{n-2}) = 2 \cdot \frac{3^{n-1} - 1}{3 - 1} = 3^{n-1} - 1 \]

Therefore:
\[ a_n = 5 + (3^{n-1} - 1) = 4 + 3^{n-1} \]

The series sum is:
\[ S_n = \sum_{n=1}^{n} (4 + 3^{n-1}) = 4n + \sum_{n=1}^{n} 3^{n-1} = 4n + \frac{3^n - 1}{2} \]

\[ S_n = \frac{8n + 3^n - 1}{2} \]
In simple words: Find what changes between consecutive terms. If those changes form a geometric sequence, use the GP sum formula to build the general term, then add all terms.

Exam Tip: When differences follow a geometric pattern with ratio > 1, the general term will contain an exponential - always separate it from the constant part when summing.

Question 24. If \( S_k = \frac{1+2+3+\cdots k}{k} \), prove that \( \left(S_1^2 + S_2^2 + \cdots S_n^2\right) = \frac{n}{24}(2n^2 + 9n + 13) \)
Answer: We are given that \( S_k = \frac{1+2+3+\cdots k}{k} \).

We need to show: \( \left(S_1^2 + S_2^2 + \cdots S_n^2\right) = \frac{n}{24}(2n^2 + 9n + 13) \)

Using the formula for the sum of the first k natural numbers:
\( S_k = \frac{k(k+1)}{2k} = \frac{k+1}{2} \)

Therefore:
\( S_k^2 = \left(\frac{k+1}{2}\right)^2 = \frac{(k+1)^2}{4} = \frac{k^2 + 2k + 1}{4} \)

The required left-hand side becomes:
\( \sum_{k=1}^{n} S_k^2 = \sum_{k=1}^{n} \frac{k^2 + 2k + 1}{4} \)

\( = \frac{1}{4}\left(\sum_{k=1}^{n} k^2 + 2\sum_{k=1}^{n} k + \sum_{k=1}^{n} 1\right) \)

\( = \frac{1}{4}\left[\frac{n(n+1)(2n+1)}{6} + 2 \cdot \frac{n(n+1)}{2} + n\right] \)

\( = \frac{1}{4}\left[n(n+1)\left(\frac{2n+1}{6}+1\right) + n\right] \)

\( = \frac{n}{4}\left[(n+1)\left(\frac{2n+7}{6}\right) + 1\right] \)

\( = \frac{n}{4}\left[\frac{(n+1)(2n+7)+6}{6}\right] \)

\( = \frac{n}{24}\left[(n+1)(2n+7) + 6\right] \)

\( = \frac{n}{24}\left[2n^2 + 7n + 2n + 7 + 6\right] \)

\( = \frac{n}{24}(2n^2 + 9n + 13) \)

Hence proved.
In simple words: Substitute the formula for \( S_k \), expand each squared term, then apply standard summation identities for natural numbers, squares, and constants to simplify and obtain the desired result.

Exam Tip: Break down compound sums by separating them into individual summations using linearity; always apply standard formulas for \( \sum k \), \( \sum k^2 \), and \( \sum 1 \) to avoid calculation errors.

Question 25. If \( S_n \) denotes the sum of the cubes of the first n natural numbers and \( s_n \) denotes the sum of the first n natural numbers, then find the value of \( \frac{\sum_{k=1}^{n} S_k}{\sum_{k=1}^{n} s_k} \)
Answer: We are given that \( S_n \) represents the sum of cubes of the first n natural numbers.

\( S_n = \sum_{k=1}^{n} k^3 = \left[\frac{n(n+1)}{2}\right]^2 \)

And \( s_n \) represents the sum of the first n natural numbers.

\( s_n = \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \)

We need to find the ratio:
\( \frac{\sum_{k=1}^{n} S_k}{\sum_{k=1}^{n} s_k} \)

\( \sum_{k=1}^{n} s_k = \sum_{k=1}^{n} \frac{k(k+1)}{2} = \frac{1}{2}\sum_{k=1}^{n} (k^2 + k) = \frac{1}{2}\left[\frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}\right] \)

\( = \frac{n(n+1)}{2}\left[\frac{2n+1}{6} + \frac{1}{2}\right] = \frac{n(n+1)}{2} \cdot \frac{2n+1+3}{6} = \frac{n(n+1)(2n+4)}{12} = \frac{n(n+1)(n+2)}{6} \)

\( \sum_{k=1}^{n} S_k = \sum_{k=1}^{n} \left[\frac{k(k+1)}{2}\right]^2 = \frac{1}{4}\sum_{k=1}^{n} k^2(k+1)^2 \)

\( = \frac{1}{4}\sum_{k=1}^{n} (k^4 + 2k^3 + k^2) \)

Using standard formulas, after simplification:
\( \sum_{k=1}^{n} S_k = \frac{n(n+1)(n+2)(3n+5)}{24} \)

Therefore:
\( \frac{\sum_{k=1}^{n} S_k}{\sum_{k=1}^{n} s_k} = \frac{\frac{n(n+1)(n+2)(3n+5)}{24}}{\frac{n(n+1)(n+2)}{6}} = \frac{n(n+1)(n+2)(3n+5)}{24} \cdot \frac{6}{n(n+1)(n+2)} = \frac{3n+5}{4} \)

In simple words: Use the formulas for cubes and natural number sums, compute each total by summing across all indices, then divide to get a simplified ratio.

Exam Tip: Always simplify summation problems by writing out the individual formula for each term, then apply telescoping or standard summation identities; cancelling common factors early saves computation time.

Exercise 13(B)

Question 1. Find the sum (2 + 4 + 6 + 8 + \( \ldots \) + 100).
Answer: We need to evaluate the sum (2 + 4 + 6 + 8 + \( \ldots \) 100).

Taking out a common factor of 2 from every term gives us:
\( 2(1 + 2 + 3 + 4 + \ldots 50) \)

Now we require the sum of the first 50 natural numbers.

Using the formula \( \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \):

Sum of the first 50 natural numbers \( = \frac{50(51)}{2} = \frac{2550}{2} = 1275 \)

Therefore:
\( (2 + 4 + 6 + 8 + \ldots 100) = 2 \times 1275 = 2550 \)

In simple words: Pull out the common factor 2, leaving you with the sum of integers 1 through 50, apply the natural numbers formula, then multiply the result by 2.

Exam Tip: Always look for common factors in series — factoring them out reduces the problem to a standard formula, making arithmetic faster and less error-prone.

Question 2. Find the sum (41 + 42 + 43 + \( \ldots \) + 100).
Answer: We need to evaluate the sum (41 + 42 + 43 + \( \ldots \) + 100).

This can be rewritten as:
\( (41 + 42 + 43 + \ldots + 100) = \) Sum of integers from 1 to 100 - Sum of integers from 1 to 40

Using the formula \( \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \):

Sum of integers from 1 to 100 \( = \frac{100(101)}{2} = \frac{10100}{2} = 5050 \)

Sum of integers from 1 to 40 \( = \frac{40(41)}{2} = \frac{1640}{2} = 820 \)

Therefore:
\( (41 + 42 + 43 + \ldots + 100) = 5050 - 820 = 4230 \)

In simple words: Subtract the sum up to 40 from the sum up to 100 to isolate the terms between 41 and 100.

Exam Tip: For sums of consecutive integers in a gap, always use the "total minus prefix" method — compute the full sum, subtract the lower portion, and verify your boundary endpoints.

Question 3. Find the sum \( 11^2 + 12^2 + 13^2 + \ldots 20^2 \)
Answer: We need to evaluate the sum \( 11^2 + 12^2 + 13^2 + \ldots 20^2 \)

This can be rewritten as:
\( 11^2 + 12^2 + 13^2 + \ldots 20^2 = \) Sum of squares from 1 to 20 - Sum of squares from 1 to 10

Using the formula \( \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \):

Sum of squares from 1 to 20 \( = \frac{20(21)(41)}{6} = \frac{17220}{6} = 2870 \)

Sum of squares from 1 to 10 \( = \frac{10(11)(21)}{6} = \frac{2310}{6} = 385 \)

Therefore:
\( 11^2 + 12^2 + 13^2 + \ldots 20^2 = 2870 - 385 = 2485 \)

In simple words: Take the sum of squares up to 20, subtract the sum of squares up to 10, and the difference gives you the sum you seek.

Exam Tip: For squared or cubed terms in a range, always compute the complete sum down to the boundary, then subtract the lower portion — this avoids tedious term-by-term addition.

Question 4. Find the sum \( 6^3 + 7^3 + 8^3 + 9^3 + 10^3 \)
Answer: We need to evaluate the sum \( 6^3 + 7^3 + 8^3 + 9^3 + 10^3 \)

This can be rewritten as:
\( 6^3 + 7^3 + 8^3 + 9^3 + 10^3 = \) Sum of cubes from 1 to 10 - Sum of cubes from 1 to 5

Using the formula \( \sum_{k=1}^{n} k^3 = \left[\frac{n(n+1)}{2}\right]^2 \):

Sum of cubes from 1 to 10 \( = \left[\frac{10(11)}{2}\right]^2 = (55)^2 = 3025 \)

Sum of cubes from 1 to 5 \( = \left[\frac{5(6)}{2}\right]^2 = (15)^2 = 225 \)

Therefore:
\( 6^3 + 7^3 + 8^3 + 9^3 + 10^3 = 3025 - 225 = 2800 \)

In simple words: Apply the cube sum formula for the range 1 to 10, apply it again for 1 to 5, then subtract to isolate the cubes from 6 to 10.

Exam Tip: Remember that the sum of cubes equals the square of the sum of natural numbers — this elegant identity saves computation when dealing with cubic series.

Question 5. If \( \sum_{k=1}^{n} k = 210 \), find the value of \( \sum_{k=1}^{n} k^2 \)
Answer: We are given that \( \sum_{k=1}^{n} k = 210 \)

Using the formula \( \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \):

\( \frac{n(n+1)}{2} = 210 \)

\( n(n+1) = 420 \)

\( n^2 + n - 420 = 0 \)

Factoring: \( (n - 20)(n + 21) = 0 \)

This gives \( n = 20 \) or \( n = -21 \)

Since n must be a positive integer, \( n = 20 \)

Now we find \( \sum_{k=1}^{n} k^2 \) using the formula \( \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \):

\( \sum_{k=1}^{20} k^2 = \frac{20(21)(41)}{6} = \frac{17220}{6} = 2870 \)

In simple words: Use the given sum to find what n is, then plug that value into the squares formula to get your answer.

Exam Tip: When you are given a sum equals a number, always solve for n first using the inverse formula — this is your key to unlocking follow-up calculations.

Question 6. If \( \sum_{k=1}^{n} k = 45 \), find the value of \( \sum_{k=1}^{n} k^3 \)
Answer: We are given that \( \sum_{k=1}^{n} k = 45 \)

Using the formula \( \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \):

\( \frac{n(n+1)}{2} = 45 \)

\( n(n+1) = 90 \)

\( n^2 + n - 90 = 0 \)

Factoring: \( (n - 9)(n + 10) = 0 \)

This gives \( n = 9 \) or \( n = -10 \)

Since n must be a positive integer, \( n = 9 \)

Now we find \( \sum_{k=1}^{n} k^3 \) using the formula \( \sum_{k=1}^{n} k^3 = \left[\frac{n(n+1)}{2}\right]^2 \):

\( \sum_{k=1}^{9} k^3 = \left[\frac{9(10)}{2}\right]^2 = (45)^2 = 2025 \)

In simple words: Solve for n from the given sum, then apply the cube sum formula which equals the square of the natural number sum.

Exam Tip: Notice that \( \sum k^3 = \left(\sum k\right)^2 \) - this connection makes cube problems elegant and helps verify your work.

Question 7. Find the sum of the series \( \{2^2 + 4^2 + 6^2 + \ldots + (2n)^2\} \)
Answer: We need to evaluate the sum \( \{2^2 + 4^2 + 6^2 + \ldots + (2n)^2\} \)

The general term of this series is \( (2n)^2 = 4n^2 \)

Therefore, the sum becomes:
\( S_n = \sum_{k=1}^{n} 4k^2 = 4 \sum_{k=1}^{n} k^2 \)

Using the formula \( \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \):

\( S_n = 4 \cdot \frac{n(n+1)(2n+1)}{6} = \frac{2n(n+1)(2n+1)}{3} \)

Therefore, the sum of the series is \( S_n = \frac{2n(n+1)(2n+1)}{3} \)

In simple words: Factor out the constant 4, apply the standard sum-of-squares formula, and simplify to get your final answer.

Exam Tip: When all terms share a common factor, extract it before using standard formulas — this reduces the expression you work with and minimizes arithmetic mistakes.

Question 8. Find the sum of 10 terms of the geometric series \( \sqrt{2} + \sqrt{6} + \sqrt{18} + \ldots \)
Answer: We need to find the sum of 10 terms of the given geometric progression.

The first term is \( a = \sqrt{2} \)

The second term is \( \sqrt{6} = \sqrt{2} \cdot \sqrt{3} \)

The third term is \( \sqrt{18} = \sqrt{2} \cdot \sqrt{9} = \sqrt{2} \cdot 3 = \sqrt{2} \cdot (\sqrt{3})^2 \)

The common ratio is \( r = \frac{\sqrt{6}}{\sqrt{2}} = \sqrt{3} \)

Using the formula for the sum of n terms of a GP: \( S_n = \frac{a(r^n - 1)}{r - 1} \)

For n = 10, a = \( \sqrt{2} \), r = \( \sqrt{3} \):

\( S_{10} = \frac{\sqrt{2}((\sqrt{3})^{10} - 1)}{\sqrt{3} - 1} = \frac{\sqrt{2}(3^5 - 1)}{\sqrt{3} - 1} = \frac{\sqrt{2}(243 - 1)}{\sqrt{3} - 1} = \frac{242\sqrt{2}}{\sqrt{3} - 1} \)

Rationalizing the denominator:
\( S_{10} = \frac{242\sqrt{2}(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{242\sqrt{2}(\sqrt{3} + 1)}{3 - 1} = \frac{242\sqrt{2}(\sqrt{3} + 1)}{2} = 121\sqrt{2}(\sqrt{3} + 1) \)

\( S_{10} = 121\sqrt{6} + 121\sqrt{2} \approx 467.5 \)

In simple words: Identify the first term and common ratio by examining consecutive terms, then apply the geometric series formula and rationalize if needed.

Exam Tip: When the common ratio involves surds, always rationalize the denominator at the end — this produces a cleaner final form and is expected in formal answers.

Question 9. Find the sum of n terms of the series whose rth term is \( (r + 2^r) \)
Answer: We need to find the sum of n terms of the series where the rth term is \( (r + 2^r) \)

The general term is: \( a_r = r + 2^r \)

The sum becomes:
\( S_n = \sum_{k=1}^{n} (k + 2^k) = \sum_{k=1}^{n} k + \sum_{k=1}^{n} 2^k \)

The first summation is the sum of natural numbers:
\( \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \)

The second summation is a geometric series with first term \( a = 2 \), common ratio \( r = 2 \):
\( \sum_{k=1}^{n} 2^k = \frac{2(2^n - 1)}{2 - 1} = 2(2^n - 1) = 2^{n+1} - 2 \)

Therefore:
\( S_n = \frac{n(n+1)}{2} + 2^{n+1} - 2 \)

In simple words: Split the series into two parts - one with natural numbers and one with powers of 2 - apply each formula separately, then combine.

Exam Tip: Whenever a general term is a sum of different types (linear plus exponential), always decompose it into separate sums — each part yields to a standard formula.

Question 1. Find the sum of n terms of the series whose rth term is (r + 2r).
Answer: The series has its rth term equal to (r + 2r). We need to split this into two separate summations:
\[ S_n = \sum_{k=1}^{n} k + \sum_{k=1}^{n} 2^k \]
The first sum is the sum of the first n natural numbers, which equals \( \frac{n(n+1)}{2} \).
The second sum is a geometric progression with first term a = 2 and common ratio r = 2. Using the GP formula:
\[ \sum_{k=1}^{n} 2^k = 2(2^n - 1) \]
Combining both results:
\[ S_n = \frac{n(n+1)}{2} + 2(2^n - 1) \]
Simplifying:
\[ S_n = \frac{n^2 + n + 4(2^n) - 4}{2} = \frac{n^2 + n - 4 + 2^{n+2}}{2} \]
In simple words: Break the series into two parts - one where you add up natural numbers from 1 to n, and another where you add powers of 2. Calculate each part using its formula, then combine them to get the final answer.

Exam Tip: Always split mixed series into standard forms (arithmetic sums, geometric progressions) - this makes the calculation straightforward and reduces errors.