Question

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

• A. $1-\frac{2^{101}}{4^{101}-1}$
• B. $1-\frac{2^{100}}{4^{100}-1}$
• C. $1+\frac{2^{100}}{4^{101}-1}$
• D. $1+\frac{2^{101}}{4^{101}-1}$

Answer 1

The correct option is A $1-\frac{2^{101}}{4^{101}-1}$

Let $S=\frac{1}{x+1}+\frac{2}{x^2+1}+\frac{2^2}{x^4+1}+\cdots +\frac{2^{100}}{x^{2^{100}}+1}$
$\Rightarrow S-\frac{1}{x-1}=(\frac{-1}{x-1}+\frac{1}{x+1})+\frac{2}{x^2+1}+\frac{2^2}{x^4+1}+\cdots +\frac{2^{100}}{x^{2^{100}}+1}$
$\Rightarrow S-\frac{1}{x-1}=(\frac{-2}{x^2-1}+\frac{2}{x^2+1})+\frac{2^2}{x^4+1}+\cdots +\frac{2^{100}}{x^{2^{100}}+1}$
$\Rightarrow S-\frac{1}{x-1}=(\frac{-2^2}{x^4-1}+\frac{2^2}{x^4+1})+\cdots +\frac{2^{100}}{x^{2^{100}}+1}$
$⋮$
$\Rightarrow S-\frac{1}{x-1}=-\frac{2^{101}}{x^{2^{101}}-1}$

Put $x=2$
$\Rightarrow S-1=-\frac{2^{101}}{2^{2^{101}}-1}$
$\Rightarrow S=1-\frac{2^{101}}{2^{2^{101}}-1}=1-\frac{2^{101}}{4^{2^{100}}-1}$