If the sum of the first 20 terms of the series (4.1)/(4+3.1^2+1^4)+(4.2)/(4+3.2^2+2^4)+(4.3)/(4+3.3^2+3^4)+(4.4)/(4+3.4^2+4^4)+ is (m)/(n) , where m and n are coprime, then m+n is equal to :-

Correct answer: C. 421 aligned & ∑_ r=1 ^ 20 4 r 4+3 r^2+r^4 & ∑_ r=1 ^ 20 4 r (r^2+r+2)(r^2-r+2) & 2 ∑_ r=1 ^ 20 ( 1 r^2-r+2 - 1 r^2+r+2 ) & 2((1)/(2)-(1)/(4)) & (1)/(4)-(1)/(8) & (1)/(8)-(1)/(14) & ((1)/(382)-(1)/(422)) & =2((1)/(2)-(1)/(422)) & =(420)/(422) & =(210)/(211) & option (3) aligned

JEE Main 20254 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 4 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Sequences and Series

JEE Main 2025 · 4 April · Evening Shift · Mathematics · Sequences and Series

If the sum of the first 20 terms of the series...

If the sum of the first 20 terms of the series

\frac{4.1}{4+3.1^2+1^4}+\frac{4.2}{4+3.2^2+2^4}+\frac{4.3}{4+3.3^2+3^4}+\frac{4.4}{4+3.4^2+4^4}+\ldots

\frac{m}{n}

, where

m

and

n

are coprime, then

m+n

is equal to :-

Options

A $423$
B $420$
C $421$
D $422$

Correct answer

C. $421$

Step-by-step solution

\begin{aligned} & \sum_{\mathrm{r}=1}^{20} \frac{4 \mathrm{r}}{4+3 \mathrm{r}^2+\mathrm{r}^4} \\ & \sum_{\mathrm{r}=1}^{20} \frac{4 \mathrm{r}}{\left(\mathrm{r}^2+\mathrm{r}+2\right)\left(\mathrm{r}^2-\mathrm{r}+2\right)} \\ & 2 \sum_{\mathrm{r}=1}^{20}\left(\frac{1}{\mathrm{r}^2-\mathrm{r}+2}-\frac{1}{\mathrm{r}^2+\mathrm{r}+2}\right) \\ & 2\left(\frac{1}{2}-\frac{1}{4}\right) \\ & \frac{1}{4}-\frac{1}{8} \\ & \frac{1}{8}-\frac{1}{14} \\ & \left(\frac{1}{382}-\frac{1}{422}\right) \\ & =2\left(\frac{1}{2}-\frac{1}{422}\right) \\ & =\frac{420}{422} \\ & =\frac{210}{211} \\ & \text{ option }(3)\end{aligned}

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