Finding a general formula for summation

Question

I am trying to find a general summation for $$\sum_{k=1}^{n} \frac{(k-1)2^k}{k(k+1)}$$ I tried to expand it and look for some pattern.
$$\frac{(1-1)2^1}{1(1+1)} + \frac{(2-1)2^2}{2(2+1)}+...+\frac{((k-1)-1)2^{k-1}}{(k-1)((k-1)+1} + \frac{(k-1)2^k}{k(k+1)}$$ If it can be solved through telescoping, I couldn't see how.

Answer 1

HINT

We have

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

and therefore

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

Answer 2

Here's the whole solution. I couldn't have figured it out without the hint that $$\frac{(k−1)2^{k}}{k(k+1)}=\frac{2^{k+1}}{k+1}-\frac{2^k}{k}$$ From there the summation can be expanded such that most of the terms factor out and all that's left is $$-2+\frac{2^{k+1}}{k+1}$$