Question

The sum of 10 terms of the series \( \frac{3}{1^2 \times 2^2} + \frac{5}{2^2 \times 3^2} + \frac{7}{3^2 \times 4^2} + \dots \) is :

• A. 1
• B. \( \frac{99}{100} \)
• C. \( \frac{120}{121} \)
• D. \( \frac{143}{144} \)

Hint:

In rational series where the numerator is the difference of factors in the denominator, the sum of \( N \) terms is almost always \( \text{First Term Part} - \text{Last Term Part} \).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We find the general term \( T_n \) and express it as a difference of two terms (Method of Differences or Telescoping Series) to calculate the partial sum.
Step 2: Detailed Explanation:
1. The \( n^{th} \) term is \( T_n = \frac{2n + 1}{n^2(n + 1)^2} \).
2. Observe the relationship between the numerator and denominator:
\[ (n + 1)^2 - n^2 = (n^2 + 2n + 1) - n^2 = 2n + 1 \]
3. Rewrite \( T_n \):
\[ T_n = \frac{(n + 1)^2 - n^2}{n^2(n + 1)^2} = \frac{1}{n^2} - \frac{1}{(n + 1)^2} \]
4. Sum of first 10 terms:
\[ S_{10} = \sum_{n=1}^{10} \left( \frac{1}{n^2} - \frac{1}{(n + 1)^2} \right) \]
\[ S_{10} = \left( \frac{1}{1^2} - \frac{1}{2^2} \right) + \left( \frac{1}{2^2} - \frac{1}{3^2} \right) + \dots + \left( \frac{1}{10^2} - \frac{1}{11^2} \right) \]
Most terms cancel out (telescoping property):
\[ S_{10} = 1 - \frac{1}{11^2} = 1 - \frac{1}{121} = \frac{120}{121} \]
Step 3: Final Answer:
The sum is \( \frac{120}{121} \).