Question

The sum of $\frac{1}{x+1}+\frac{2}{x^2+1}+\frac{4}{x^4+1}+....+\frac{2^n}{x^{2n}+1}$ is

• A. $2^n\frac{1}{1-x^{n+1}}-\frac{1}{(1-x)}$
• B. $2^n\frac{1}{1-x^{2n+1}}$
• C. $2^{n+1}\frac{1}{1-x^{2n+1}}-\frac{1}{1-x}$
• D. $2^{n+1}\frac{1}{1-x^{2n+1}}$

Answer 1

The correct option is C $2^{n+1}\frac{1}{1-x^{2n+1}}-\frac{1}{1-x}$

$S_n=\frac{1}{x+1}+\frac{2}{x^2+1}+...+\frac{2^n}{x^{2n}+1}$
$S_n+\frac{1}{1-x}$
$=\frac{1}{1-x}+\frac{1}{x+1}+\frac{2}{x^2+1}+...+\frac{2^n}{x^{2n}+1}$
$=\frac{2}{1-x^2}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+....$
$=\frac{4}{1-x^4}+\frac{4}{1+x^4}+...+\frac{2^n}{x^{2n}+1}$
Subsequently,
$S_n+\frac{1}{1-x}$
$=\frac{2^n}{1-x^{2n}}+\frac{2^n}{x^{2n}+1}$
$=\frac{2^{n+1}}{1-x^{2n+1}}$
$\therefore S_n=\frac{2^{n+1}}{1-x^{2n+1}}-\frac{1}{1-x}$