Question

Prove that mathematical induction that
$\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+...+\frac{2^n}{1+x^{2^n}}$
$=\frac{1}{1-}\frac{2^{n+i}}{1+x^{2^{n+i}}}$

Answer 1

P(0) = $\frac{1}{1+x},\frac{1}{1-x}+\frac{2}{1+x^2}=\frac{1}{1-x}-\frac{2}{x^2-1}$
$=\frac{x+1-2}{(x^2-1)}=\frac{1}{1+x}$
Thus P(0) holds goods. Assume P(n)
P(n + 1) = P(n) + $\frac{2^{n+1}}{1+x^{2^{n+1}}}$
$=\frac{1}{1+x}+\frac{2^{n+1}}{1-x^{2^{n+1}}}+\frac{2^{n+1}}{1+x^{2^{n+1}}}$
$=\frac{1}{1+x}+\frac{2^{n+1}(1+1)}{1-(x^{2^{n+1}}{)}^2}=\frac{1}{x-1}+\frac{2^{n+2}}{1-x^{2^{n+2}}}$