Question

Obtain the sum of $\frac{1}{x+1}+\frac{2}{x^2+1}+\frac{4}{x^4+1}+....+\frac{2^n}{x^{2n}+1}$

• A. $\frac{2^{n+1}}{1-x^{2n+1}}-\frac{1}{1-x}$
• B. $\frac{2^{n+1}}{1-x^{2n+1}}+\frac{1}{1+x^2}$
• C. $\frac{2^{n+1}}{1-x^{n+1}}-\frac{1}{1+x^2}$
• D. None of these

Answer 1

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

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