Use induction to prove $\sum_{k=1}^n\frac{k}{2^k}=2-\frac{n+2}{2^n}$ 
Use induction on $n\in\Bbb N$ to prove that $$\sum_{k=1}^n\frac{k}{2^k}=2-\frac{n+2}{2^n}\;.$$

I have got as far as to the induction step where I have:
$$S(n+1)= 2-\frac{n+3}{2^{n+1}}$$ and this should be equal to
$$S(n) +\frac{n+1}{2^{n+1}} = 2-\frac{n+2}{2^n}+\frac{n+1}{2^{n+1}}$$
Now I am kinda stuck.. all I end up with is ;
$$ ... = 2-\frac{3n+5}{2^{n+1}}\;.$$
Would someone help me out?

When making a common denominator, don't forget to distribute the negative sign!
$$
\frac{-(n + 2)}{2^n} + \frac{n+1}{2^{n+1}}
= \frac{-2(n + 2) + (n + 1)}{2^{n+1}}
= \frac{(-2n - 4) + (n + 1)}{2^{n+1}}
= \frac{-n - 3}{2^{n+1}}
= -\frac{n + 3}{2^{n+1}}
$$
