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.
 A: 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.
A: In $P2$ you have made a misktake. It will be, $$\dfrac{2^{k+1}-k\color{red}{-}2}{2^k}+\dfrac{k+1}{2^{k+1}}=\dfrac{2^{k+2}-k\color{red}{-}3}{2^{k+1}}$$
Now if you simplify LHS then,
 $$\dfrac{2^{k+2}-2k-4+k+1}{2^{k+1}}=\dfrac{2^{k+2}-k-3}{2^{k+1}}=2-\dfrac{k+3}{2^{k+1}}$$and you are done.
