Mathematics · Sequence and Series

JEE Main 2025 — 2 April, Evening Shift — Question 42

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 $\frac{m}{n}, \quad$ where

$\operatorname{gcd}(m, n)$ , then $m+n$ is equal to $\qquad$ .

Answer: 441

Numerical answer — enter this value.

Step-by-step solution

$\frac{4.1}{1+4.1^{4}}+\frac{4.2}{1+4.2^{4}}+\frac{4.3}{1+4.3^{4}}+\ldots .$ .

\begin{aligned} & T_{r}=\frac{4 r}{1+4 r^{4}}=\frac{4 r}{4 r^{4}+4 r^{2}+1-4 r^{2}} \\& =\frac{4 r}{\left(2 r^{2}+1\right)^{2}-(2 r)^{2}} \\& T_{r}=\frac{4 r}{\left(2 r^{2}-2 r+1\right)\left(2 r^{2}+2 r+1\right)} \end{aligned}

\begin{aligned} & T_{r}=\frac{\left(2 r^{2}+2 r+1\right)-\left(2 r^{2}-2 r+1\right)}{\left(2 r^{2}-2 r+1\right)\left(2 r^{2}+2 r+1\right)} \\& T_{r}=\left(\frac{1}{r^{2}+(r-1)^{2}}-\frac{1}{r^{2}+(r+1)^{2}}\right) \\& \sum_{r=1}^{10} T_{r}=\left(\frac{1}{0^{2}+1^{2}}-\frac{1}{1^{2}+2^{2}}+\frac{1}{1^{2}+2^{2}}-\frac{1}{2^{2}+3^{2}}+\ldots\right. \\& \qquad \frac{1}{9^{2}+10^{2}}-\frac{1}{10^{2}+11^{2}} \\& =1-\frac{1}{221} \\& =\frac{220}{221} \end{aligned}

\therefore m+n=220+221

=441

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 2 April, Evening Shift
Subject: Mathematics
Chapter: Sequence and Series
Topic: Telescopic Summation

The number of terms of an A.P. is even, the sum of all the odd terms is 24 , the sum of all the even terms is 30 and the last term exceeds…

Answer: 441

Step-by-step solution

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