Prove that $\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3} +...+ \frac{n}{2^n} = 2 - \frac{n+2}{2^n} $ I need help with this exercise from the book What is mathematics? An Elementary Approach to Ideas and Methods. Basically I need to proove:
$$\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3} +...+ \frac{n}{2^n} = 2 - \frac{n+2}{2^n} $$
$i)$ Particular cases
$ Q(1) = \frac{1}{2} ✓ $ ---------- $ P(1) = 2 - \frac{1+2}{2^1}  = \frac{1}{2} ✓ $
$+ = 1 ✓ (=Q(1)+Q(2))$ 
$ Q(2) = \frac{2}{4}(=\frac{1}{2})$----$P(2) = 2 - \frac{2+2}{2^2} = 1 $ ✓
$ii)$ Hypothesis
$$F(k)=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3} +...+ \frac{k}{2^k} = 2 - \frac{k+2}{2^k} $$
$iii)$ Proof
$P1 | F(k+1) = ( 2 -\frac{k+2}{2^k} ) + ( \frac{k+1}{2^{k+1}}) = 2 - \frac{k+3}{2^{k+1}}  $
$P2 | \frac{2^{k+1}-k-2}{2^k} + \frac{k+1}{2^{k+1}} = \frac{2^{k+2}-k-3}{2^{k+1}} $
$P3 | \frac{2^{2k+2}-2^{k+1}K-2^{k+2}+2^kk+2k}{2^{2k+1}} = \frac{2^{k+2}-k-3}{2^{k+1}} $
$P4 | \frac{2(2^{k+1}-2^kk-2^{k+1}-2^{k-1}k+k)}{2^{k+1}} = \frac{2^{k+2}-k-3}{2^{k+1}} $
I get stuck here; if you could help me, that would be really kind.
Thanks in advance.

You don't need any induction for this. This Series is simple AGP. Procedure for solving these series is assume 
$$s=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}.....\frac{n-1}{2^{n-1}}+\frac{n}{2^n}$$ Now multiply whole series by $\frac{1}{2}$ and write the whole series as one term shifted $$\frac{s}{2}= \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{2^2}+\frac{2}{2^3}+\frac{3}{2^4}......\frac{n-1}{2^{n}}+\frac{n}{2^{n+1}}$$ Second series should be subtracted from first in such a way that terms which has same powers of $2$ comes togethar. Now subtract series $$\frac{s}{2}=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}......\frac{1}{2^{n}}-\frac{n}{2^{n+1}}$$ First n terms are of GP and can be written as $(1-\frac{1}{2^n})$, so $\frac{s}{2}=1-\frac{n+2}{2^{n+1}}$ . Now multiply with $2$ so as to get the correct expression for s $$s=2-\frac{n+2}{2^{n}}$$ Hence proved without induction.
