Question

Let $S=\sum _{k=0}^{\infty }\frac{2^k}{7^{2^k}+1}$ then $\frac{1}{S}$ is equal to

Answer 1

Answer: 6

$S=\sum _{k=0}^{\infty }\left(\frac{2^k}{7^{2^k}+1}\right)$
$S=\frac{2^0}{7^1+1}+\frac{2^1}{7^2+1}+\frac{2^2}{7^4+1}+\frac{2^3}{7^8+1}$
$\left(\frac{1}{7-1}-\frac{2}{7^2-1}\right)+\left(\frac{2}{7^2-1}-\frac{2^2}{7^4-1}\right)+\left(\frac{2^2}{7^4-1}-\frac{2^3}{7^8-1}\right)+\dots$
$S=\frac{1}{6}\Rightarrow \frac{1}{5}=6$