Evaluate $\sum_{r=1}^{n}[\sum_{k=1}^{r}k][\log_{1/2}\sqrt{4x-x^2}]^r$

Question

$$ {\rm Evaluate} \sum_{r=1}^{n}\Big(\sum_{k=1}^{r}k\Big)\big(\log_{1/2}\sqrt{4x-x^2}\big)^{r} $$

First step is probably to make it into $S=\sum_{r=1}^{n}[\frac{r(r+1)}{2}]A^r$ where $A=\log_{1/2}\sqrt{4x-x^2}$

then, I tried to do with 2 methods but I am struck in both of them. 1st way is using A.G.P where I get $(1-A)S=\sum_{r=1}^{n}rA^r$ and $(1+A)S=\sum_{r=1}^{n}r^{2}A^{r}$ not sure what to do now and another try was to use binomial theorem but the result i.e $S=\sum_{r=1}^{n}\mathrm{C}_{2}^{r+1}A^r$ I cannot think a way to convert to do so.

Is there a way to do a method to do this type of question under 2 mins? i.e Method which will not be too lengthy.

Answer 1

Hint

As you wrote, we need to consider that $$2S=\sum_{r=1}^n r(r+1)A^r$$ Now, the trick is to use $$r(r+1)=r(r-1)+2r$$ $$2S=\sum_{r=1}^n r(r-1)A^r+2\sum_{r=1}^n rA^r=A^2\sum_{r=1}^n r(r-1)A^{r-2}+2A\sum_{r=1}^n rA^{r-1}$$ $$2S=A^2 \Bigg[\sum_{r=1}^n A^{r}\Bigg]''+2A\Bigg[\sum_{r=1}^n A^{r}\Bigg]'$$