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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.

Martin Sleziak
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Daniel Bonilla Jaramillo
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  • In $P1$, you should leave $2$ as itself since it must appear in the answer. It will probably make the fraction more manageable. – JasonM Jun 12 '16 at 06:09
  • Multiply the expression by $2^n$, and then follow http://math.stackexchange.com/questions/87030/proving-sum-limits-i-0n-i-2i-1-n1-2n-1-by-induction – Mc Cheng Jun 12 '16 at 06:09
  • Just focus on turning $-\frac{k+2}{2^k}+\frac{k+1}{2^{k+1}}$ into $-\frac{k+3}{2^{k+1}}$ – JasonM Jun 12 '16 at 06:12
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    Do you know how to sum the geometric series $1+\frac{1}{2}+\frac{1}{2^2}+\cdots+\frac{1}{2^n}$? – John Wayland Bales Jun 12 '16 at 06:47

2 Answers2

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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.

Vineet Mangal
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  • +1 Yes, I prefer that approach, although it will probably irritate the OP! – almagest Jun 12 '16 at 09:19
  • This is the general approach for these types of series. BTW who is OP ? – Vineet Mangal Jun 12 '16 at 09:36
  • Yes, OP is a bizarre abbreviation, but widely used on this site. It stands for Original Poster, meaning the person who asked the question. But I doubt if 5% of the people who use it know that. – almagest Jun 12 '16 at 09:39
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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.