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 ______.

Correct answer: 0 For the series (1)/(5) + (2)/(65) + (3)/(325) + (4)/(1025) + (5)/(2501) + T _ n = n 4 n ^ 4 + 1 = n (2 n ^ 2 + 1) ^ 2 - ( 2n ) ^ 2 = n (2 n ^ 2 + 2n + 1)(2 n ^ 2 - 2n + 1) = (1)/(4)[ 1 2 n ^ 2 - 2n + 1 - 1 2 n ^ 2 + 2n + 1 ] Now S _ 10 = ∑ _ n = 1 ^ 10 T _ n = ∑ _ n = 1 ^ 10 (1)/(4)[ 1 2 n ^ 2 - 2n + 1 - 1 2 n ^ 2 + 2n + 1 ] = (1)/(4)[(1)/(1) - (1)/(5) + (1)/(5) - (1)/(13) + (1)/(13) + (1)/(181) - (1)/(221)] = (1)/(4)[1 - (1)/(221)] = (55)/(221) = (m)/(n) Hence, m + n = 55 + 221 = 276

JEE Main 202227 JuneMorning Shift

Exam: JEE Main
Year: 2022
Date: 27 June
Shift: Morning Shift
Subject: Mathematics
Chapter: Sequences and Series

JEE Main 2022 · 27 June · Morning Shift · Mathematics · Sequences and Series

If the sum of the first ten terms of the series (1)/(5) + (2)/(65)...

If the sum of the first ten terms of the series

\dfrac{1}{5} + \dfrac{2}{65} + \dfrac{3}{325} + \dfrac{4}{1025} + \dfrac{5}{2501} + \ldots .

\dfrac{m}{n}

, where

m

and

n

are co-prime numbers, then

m + n

is equal to ______.

Correct answer

Step-by-step solution

For the series $\dfrac{1}{5} + \dfrac{2}{65} + \dfrac{3}{325} + \dfrac{4}{1025} + \dfrac{5}{2501} + \ldots \ldots$

${T}_{n} = \dfrac{n}{4{n}^{4} + 1}$

$= \dfrac{n}{{\left(2{n}^{2} + 1\right)}^{2} - ( 2n{)}^{2}} = \dfrac{n}{\left(2{n}^{2} + 2n + 1\right)\left(2{n}^{2} - 2n + 1\right)}$

$= \dfrac{1}{4}\left[\dfrac{1}{2{n}^{2} - 2n + 1} - \dfrac{1}{2{n}^{2} + 2n + 1}\right]$

Now ${S}_{10} = \mathop{∑}_{n = 1}^{10}{T}_{n} = \mathop{∑}_{n = 1}^{10}\dfrac{1}{4}\left[\dfrac{1}{2{n}^{2} - 2n + 1} - \dfrac{1}{2{n}^{2} + 2n + 1}\right]$

$= \dfrac{1}{4}\left[\dfrac{1}{1} - \dfrac{1}{5} + \dfrac{1}{5} - \dfrac{1}{13} + \dfrac{1}{13} \ldots \ldots \ldots + \dfrac{1}{181} - \dfrac{1}{221}\right]$

$= \dfrac{1}{4}\left[1 - \dfrac{1}{221}\right] = \dfrac{55}{221} = \dfrac{m}{n}$

Hence, $m + n = 55 + 221 = 276$

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