If the sum of the first 10 terms of the series (4.1)/(1+4.1^4)+(4.2)/(1+4.2^4)+(4.3)/(1+4.3^4)+ is (m)/(n) , where gcd(m, n)=1 , then m+n is equal to______

Correct answer: 0 aligned & T_ r = 4 . r 1+4 . r^4 & ~T_ r = 4 . r (2 r^2+2 r+1)(2 r^2-2 r+1) & T_ r = (2 r^2+2 r+1)-(2 r^2-2 r+1) (2 r^2+2 r+1)(2 r^2-2 r+1) aligned aligned & T_ r = 1 2 r^2-2 r+1 - 1 2 r^2+2 r+1 & ~T_1=(1)/(1)-(1)/(5) & ~T_2=(1)/(5)-(1)/(13) aligned aligned & T_ 10 =(1)/(181)-(1)/(221) & ~S_ 10 =1-(1)/(221)=(220)/(221)= m n & m+n=441 aligned

JEE Main 20252 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 2 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Sequences and Series

JEE Main 2025 · 2 April · Evening Shift · Mathematics · Sequences and Series

If the sum of the first 10 terms of the 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}+\ldots

\frac{m}{n}

, where

\operatorname{gcd}(m, n)=1

, then

m+n

is equal to______

Correct answer

Step-by-step solution

\begin{aligned} & \mathrm{T}_{\mathrm{r}}=\frac{4 . \mathrm{r}}{1+4 . \mathrm{r}^4} \\ & \mathrm{~T}_{\mathrm{r}}=\frac{4 . \mathrm{r}}{\left(2 \mathrm{r}^2+2 \mathrm{r}+1\right)\left(2 \mathrm{r}^2-2 \mathrm{r}+1\right)} \\ & \mathrm{T}_{\mathrm{r}}=\frac{\left(2 \mathrm{r}^2+2 \mathrm{r}+1\right)-\left(2 \mathrm{r}^2-2 \mathrm{r}+1\right)}{\left(2 \mathrm{r}^2+2 \mathrm{r}+1\right)\left(2 \mathrm{r}^2-2 \mathrm{r}+1\right)}\end{aligned}

\begin{aligned} & \mathrm{T}_{\mathrm{r}}=\frac{1}{2 \mathrm{r}^2-2 \mathrm{r}+1}-\frac{1}{2 \mathrm{r}^2+2 \mathrm{r}+1} \\ & \mathrm{~T}_1=\frac{1}{1}-\frac{1}{5} \\ & \mathrm{~T}_2=\frac{1}{5}-\frac{1}{13}\end{aligned}

\begin{aligned} & \mathrm{T}_{10}=\frac{1}{181}-\frac{1}{221} \\ & \mathrm{~S}_{10}=1-\frac{1}{221}=\frac{220}{221}=\frac{\mathrm{m}}{\mathrm{n}} \\ & \mathrm{m}+\mathrm{n}=441\end{aligned}

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