Answer
$x-x\ln(1-x)+\ln (1-x) $
Work Step by Step
We are given the series $S=\Sigma_{k=2}^{\infty} \dbinom{x^k}{k(k-1)}$
Our aim is to find the function which represent the given series.
Now, we have:
$S=\Sigma_{k=2}^{\infty} \dbinom{x^k}{k(k-1)}\\=\Sigma_{k=2}^{\infty} (\dfrac{x^k}{k-1}-\dfrac{x^k}{k})\\=x\Sigma_{k=2}^{\infty} (\dfrac{x^{k-1}}{k-1})-(\Sigma_{k=1}^{\infty}\dfrac{x^k}{k}-\dfrac{x^1}{1})\\=x \Sigma_{k=1}^{\infty}\dfrac{x^k}{k}-(\Sigma_{k=1}^{\infty}\dfrac{x^k}{k}-x)\\=x(-\ln(1-x))-[(-\ln (1-x))-x]\\=x-x\ln(1-x)+\ln (1-x) $
