The sum of the series $\frac{1}{x+1}+\frac{2}{x^{2}+1}+\frac{2^{2}}{x^{4}+1}+\ldots .+\frac{2^{100}}{x^{2^{100}}+1}$ when $x=2$ is:

(B) $1-\frac{2^{101}}{4^{101}-1}$

(A) $1-\frac{2^{100}}{4^{100}-1}$

(C) $1+\frac{2^{100}}{4^{101}-1}$

(D) $1+\frac{2^{101}}{4^{101}-1}$

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Sequences and Series

The sum of the series

\frac{1}{x + 1} + \frac{2}{x ^{2} + 1} + \frac{2 ^{2}}{x ^{4} + 1} + \dots . + \frac{2 ^{100}}{x ^{2^{100}} + 1}

when

x = 2

is:

Answer: (B)

For each

k

, putting

a = x^{2^{k}}

so that

a^{2} = x^{2^{k + 1}}

\frac{2 ^{k}}{x ^{2^{k}} - 1} - \frac{2 ^{k + 1}}{x ^{2^{k + 1}} - 1} = \frac{2 ^{k} ( a + 1 ) - 2 ^{k + 1}}{a ^{2} - 1} = \frac{2 ^{k} ( a - 1 )}{( a - 1 ) ( a + 1 )} = \frac{2 ^{k}}{x ^{2^{k}} + 1} .

Every term of the series is of this form, with

k

running from

0

100

, so the sum telescopes and only the first and last pieces survive:

S = \frac{1}{x - 1} - \frac{2 ^{101}}{x ^{2^{101}} - 1} .

Putting

x = 2

gives

S = 1 - \frac{2 ^{101}}{2 ^{2^{101}} - 1} .

JEE Main · 2021 · 26 Aug 2021 · Shift 1 · Single correct option