JEE MainMathematics

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Summation of a Telescoping Series Using Sophie Germain's Identity

Q: How do we spot that the denominator sequence is 4n^4 + 1?

Notice the sequence: 5 = 4(1) + 1, 65 = 4(16) + 1, 325 = 4(81) + 1, 1025 = 4(256) + 1. The numbers subtracted by 1 are 4, 64, 324, 1024, which are 4 × 1^4, 4 × 2^4, 4 × 3^4, 4 × 4^4.

If the sum of the first ten terms of the series 1/5+2/65+3/325+4/1025+5/2501+…. is m/n, where m and n are co-prime numbers, then m+n is equal to

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Question type: Numerical
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 22 September 2026

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How do we spot that the denominator sequence is

4 n^{4} + 1

Notice the sequence: $5 = 4 (1) + 1$ , $65 = 4 (16) + 1$ , $325 = 4 (81) + 1$ , $1025 = 4 (256) + 1$ . The numbers subtracted by 1 are $4, 64, 324, 1024$ , which are $4 \times 1^{4}, 4 \times 2^{4}, 4 \times 3^{4}, 4 \times 4^{4}$ .

Answer

Accepted answer: 276

The sum of the first ten terms is $\frac{55}{221}$ , giving $m + n = 276$ .

Why each option works or fails

Step-by-step solution

Given: The series is $\frac{1}{5} + \frac{2}{65} + \frac{3}{325} + \frac{4}{1025} + \frac{5}{2501} + \dots$ and its sum of the first 10 terms is $\frac{m}{n}$ where $\gcd (m, n) = 1$ .
Goal: Find the value of $m + n$ .
Approach: Identify the general term $T_{n}$ . The numerators are $n$ . The denominators are $5, 65, 325, 1025, 2501$ , which match $4 n^{4} + 1$ . Factor the denominator using Sophie Germain's identity: $4 n^{4} + 1 = (2 n^{2} + 1)^{2} - (2 n)^{2} = (2 n^{2} - 2 n + 1) (2 n^{2} + 2 n + 1)$ . Express $T_{n}$ as partial fractions and telescope the sum.
Execute: Write $T_{n} = \frac{n}{4 n^{4} + 1} = \frac{1}{4} [\frac{(2 n^{2} + 2 n + 1) - (2 n^{2} - 2 n + 1)}{(2 n^{2} - 2 n + 1) (2 n^{2} + 2 n + 1)}] = \frac{1}{4} [\frac{1}{2 n^{2} - 2 n + 1} - \frac{1}{2 n^{2} + 2 n + 1}]$ . Let $f (n) = 2 n^{2} - 2 n + 1$ . Then $f (n + 1) = 2 (n + 1)^{2} - 2 (n + 1) + 1 = 2 n^{2} + 2 n + 1$ .
Execute: Sum $T_{n}$ from $n = 1$ to $10$ : $S_{10} = \frac{1}{4} \sum_{n = 1}^{10} [\frac{1}{f (n)} - \frac{1}{f (n + 1)}] = \frac{1}{4} [\frac{1}{f (1)} - \frac{1}{f (11)}]$ . Since $f (1) = 2 (1)^{2} - 2 (1) + 1 = 1$ and $f (11) = 2 (10)^{2} + 2 (10) + 1 = 221$ , we have $S_{10} = \frac{1}{4} (1 - \frac{1}{221}) = \frac{1}{4} \times \frac{220}{221} = \frac{55}{221}$ .
Execute: Check if $55$ and $221$ are co-prime: $55 = 5 \times 11$ . $n = 1$ . Thus $\gcd (55, 221) = 1$ , so $m = 55$ and $n = 221$ . Hence $m + n = 55 + 221 = 276$ .
Verify: Check for $n = 1$ : $S_{1} = \frac{1}{4} (1 - \frac{1}{5}) = \frac{1}{5}$ , matching the first term. For $n = 2$ : $S_{2} = \frac{1}{5} + \frac{2}{65} = \frac{15}{65} = \frac{3}{13}$ ; from the formula $\frac{1}{4} (1 - \frac{1}{13}) = \frac{3}{13}$ . Formula holds consistently.

Summation of a Telescoping Series Using Sophie Germain's Identity

What feels right?

The sum of the first ten terms is $\frac{55}{221}$ , giving $m + n = 276$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

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