need to find sum of $\sum_{k=1}^n \frac{(k-1)2^k}{k(k+1)}$

Question

need to find sum of

$$\sum_{k=1}^n \frac{(k-1)2^k}{k(k+1)}$$

this is much harder as I see. Or maybe it's my eyes, because I'm new in studying sums.

I tried this one using differences , factorial powers, didn't help.

Answer 1

Try to get a telescoping series as follows, $$\sum_{k=1}^n \frac{(k-1)2^k}{k(k+1)}=\sum_{k=1}^n \frac{(\color{red}2k-(k+1))2^k}{k(k+1)}=\sum_{k=1}^n \frac{2^{k\color{red}{+1}}}{k+1}-\sum_{k=1}^n \frac{2^k}{k}=\frac{2^{n+1}}{n+1}-\frac{2^1}{1}$$

Answer 2

Simple telescopic series $$ \sum_{k = 1}^{n} \frac{(k-1)2^k}{k(k + 1)} = \sum_{k = 1}^{n} \bigg( \frac{2^{k + 1}}{k + 1} - \frac{2^{k}}{k} \bigg) = \frac{2^{n + 1}}{n + 1} - {2} $$