Given:

$\displaystyle f(x)=\sum^{\infty}_{n=1}\frac{2^n}{n}x^n$

$\displaystyle x\in]-1,1[$

Find a differential equation $\displaystyle f'(x)$, such that when you insert $\displaystyle x$ in the equation, you get $\displaystyle f'(x)$.

Answer:

$\displaystyle f(x)=\frac{2}{1-2x}$

I'm not sure I entirely understand this question or how to start.