Closed Form of Series $ \frac {2}{3+1} + \frac {2^2}{3^2+1} + \cdots + \frac {2^{n+1}}{3^{2^n}+1} $.

Question

Problem:

A closed form for the sum $$ S = \frac {2}{3+1} + \frac {2^2}{3^2+1} + \cdots + \frac {2^{n+1}}{3^{2^n}+1} $$ is $\Large1 - \frac{a^{n+b}}{3^{2^{n+c}}-1},$ where $a, b,$ and $c$ are integers. Find $a+b+c.$

How would I begin to do this? Any easy theorem/trick?

Answer 1

HINT:

$$\dfrac1{3^{2^m}-1}-\dfrac1{3^{2^m}+1}=\dfrac2{3^{2^{m+1}}-1}$$

$$\dfrac2{3-1}-S=\dfrac2{3-1}-\dfrac2{3+1}-\left(\dfrac{2^2}{3^2+1}+\dfrac{2^3}{3^4+1}+\cdots+\dfrac{2^n}{3^{2^n}+1}\right)$$

$$=\dfrac{2^2}{3^2-1}-\dfrac{2^2}{3^2+1}-\left(\dfrac{2^3}{3^4+1}+\cdots+\dfrac{2^n}{3^{2^n}+1}\right)=\cdots$$