If the sum of the first 10 terms of the series $\frac{4.1}{1+4.1^{4}}+\frac{4.2}{1+4.2^{4}}+\frac{4.3}{1+4.3^{4}}+\ldots$ is $\frac{m}{n}$, where $\gcd(m, n)=1$, then $m+n$ is equal to ______.

JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Sequences and Series

If the sum of the first 10 terms of the series

\frac{4.1}{1 + 4. 1 ^{4}} + \frac{4.2}{1 + 4. 2 ^{4}} + \frac{4.3}{1 + 4. 3 ^{4}} + \dots

\frac{m}{n}

, where

g cd (m, n) = 1

, then

m + n

is equal to ______.

Answer: 441

The general term, with the numerators and denominators read as

4 r

and

1 + 4 r^{4}

, is

T_{r} = \frac{4 r}{1 + 4 r ^{4}}

. By the Sophie Germain identity,

1 + 4 r^{4} = (2 r^{2} - 2 r + 1) (2 r^{2} + 2 r + 1)

, and since

4 r = (2 r^{2} + 2 r + 1) - (2 r^{2} - 2 r + 1)

, the term splits as

T_{r} = \frac{1}{2 r ^{2} - 2 r + 1} - \frac{1}{2 r ^{2} + 2 r + 1}

. Because

2 r^{2} + 2 r + 1 = 2 (r + 1)^{2} - 2 (r + 1) + 1

, the sum telescopes:

\sum_{r = 1}^{10} T_{r} = 1 - \frac{1}{2 ( 10 ) ^{2} + 2 ( 10 ) + 1} = 1 - \frac{1}{221} = \frac{220}{221}

. As

g cd (220, 221) = 1

, we have

m = 220

n = 221

, so

m + n = 441

JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Integer answer