Mathematics · 2021
JEE Main · 26 August 2021, Shift 1 · Q65
The sum of the series 1/(x+1)+2/(x^2+1)+2^2/(x^4+1)+…+(2^100)/(x^2^100 +1) when x=2 is:
The sum of the series $\displaystyle \frac{1}{x+1}+\frac{2}{x^2+1}+\frac{2^2}{x^4+1}+\ldots+\frac{2^{100}}{x^{2^{100}}+1}$ when $\displaystyle x=2$ is :
Official answer
(1)
$\displaystyle 1-\frac{2^{101}}{4^{101}-1}$
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JEE Main 2021 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.