Prove that $(x-1)\log(1 - 2 x) -2x = \sum_{n=2}^{\infty} \frac{2^n (n-1) x^{n+1}}{n^2 +n} \quad \text{for} \; 2|x|<1$ Prove that $$(x-1)\log(1 - 2 x) -2x = \sum_{n=2}^{\infty} \frac{2^n (n-1) x^{n+1}}{n^2 +n} \quad \text{for} \; 2|x|<1$$
First of all, I don't really know if by proving it means finding the function Sum of $\sum_{n=2}^{\infty} \frac{2^n (n-1) x^{n+1}}{n^2 +n}$ and concluding that $f(x)=(x-1)\log(1 - 2 x) -2x$ or to replace $\log(1 - 2 x)$ for it's Taylor Expansion and resaulting in $ \sum_{n=2}^{\infty} \frac{2^n (n-1) x^{n+1}}{n^2 +n}$. I've tried both ways, starting from the Taylor Expansion of $\log(1 - 2 x)=\sum_{n=1}^{\infty} -\frac{2^n x^{n}}{n}$ but I've failed both ways. Any hints on how to prove this? Thanks in advance.
Edit: What I did
$$\log(1 - 2 x)=\sum_{n=1}^{\infty} -\frac{2^n x^{n}}{n} = -2x + \sum_{n=2}^{\infty} -\frac{2^n x^{n}}{n} \\ \rightarrow  (x-1)\log(1 - 2 x)=-2x^2 + x\sum_{n=2}^{\infty} -\frac{2^n x^{n}}{n} + 2x +\sum_{n=2}^{\infty} \frac{2^n x^n}{n} \\ \rightarrow (x-1)\log(1 - 2 x) -2x =-2x^2 + \sum_{n=2}^{\infty} -\frac{2^n x^{n+1}}{n} + \sum_{n=2}^{\infty} \frac{2^n x^n}{n} \\ = \sum_{n=1}^{\infty} -\frac{2^n x^{n+1}}{n} + \sum_{n=2}^{\infty} \frac{2^n x^n}{n} \\ = \sum_{n=1}^{\infty} -\frac{2^{n} x^{n+1}}{n} + \sum_{n=1}^{\infty} \frac{2^{n+1} x^{n+1}}{n+1} \\ = \sum_{n=1}^{\infty} \frac{-2^{n} x^{n+1} (n+1) + 2^{n+1} x^{n+1} n}{n(n+1)} \\ = \sum_{n=1}^{\infty} \frac{-2^{n} x^{n+1}[(n+1)-2n]}{n(n+1)} \\ = \sum_{n=1}^{\infty} \frac{-2^{n} x^{n+1}(1-n)}{n(n+1)} \\ = \sum_{n=1}^{\infty} \frac{2^{n} x^{n+1}(n-1)}{n(n+1)} \\ = 0 + \sum_{n=2}^{\infty} \frac{2^{n} x^{n+1}(n-1)}{n(n+1)} \\ =  \sum_{n=2}^{\infty} \frac{2^{n} x^{n+1}(n-1)}{n(n+1)}$$
Which is what I was trying to prove.

$$\log(1-2x)=-\sum_{n=1}^\infty\frac{(2x)^n}n\;,\;\;|2x|<1\iff |x|<\frac12\implies$$
$$(x-1)\log(1-2x)-2x=\sum_{n=1}^\infty\frac{2^nx^{n+1}-2^nx^n}n-2x$$
and
$$\sum_{n=1}^\infty\frac{2^n(n-1)x^{n+1}}{n^2+n}=\overbrace{\sum_{n=1}^\infty\frac{2^nx^{n+1}}{n+1}}^{:=f(x)}-\overbrace{\sum_{n=1}^\infty\frac{2^nx^{n+1}}{n^2+n}}^{:=g(x)}$$
and now
$$f'(x)=\sum_{n=1}^\infty2^nx^n\;,\;\;g'(x)=\sum_{n=1}^\infty\frac{2^nx^n}n\implies f'(x)-g'(x)=\frac1{1-2x}-1+\log(1-2x)$$
Now try to take it from here...
