- #1

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## Homework Statement

Show that

[tex]\sum_{k=0}^{n} \frac{k}{2^{k}} = 2 - \frac{n+2}{2^{n}}~; ~~~ \forall n \in \mathbb{N} \cup \{0\}[/tex]

## The Attempt at a Solution

__(1) Show that it is true for n = 0__

[tex]\frac{0}{2^{0}} = 2 - \frac{2}{2^{0}} = 0[/tex]

__(2) Show that if it is true for n = p, it is true for n = p + 1__

Assume that

[tex]\sum_{k=0}^{p} \frac{k}{2^{k}} = 2 - \frac{p+2}{2^{p}}[/tex]

Now,

[tex]\sum_{k=0}^{p} \frac{k}{2^{k}} = \frac{1}{2} + \frac{2}{4} +\frac{3}{8} + \frac{4}{16} + ... + \frac{p}{2^{p}}[/tex]

[tex]\sum_{k=0}^{p+1} \frac{k}{2^{k}} = \frac{1}{2} + \frac{2}{4} +\frac{3}{8} + \frac{4}{16} + ... + \frac{p}{2^{p}} + \frac{p+1}{2^{p+1}} = \sum_{k=0}^{p} \frac{k}{2^{k}} + \frac{p+1}{2^{p+1}} = 2 - \frac{p+2}{2^{p}} + \frac{p+1}{2^{p+1}}[/tex]

If it can be shown that

[tex]2 - \frac{p+2}{2^{p}} + \frac{p+1}{2^{p+1}} = 2 - \frac{p+3}{2^{p+1}}[/tex]

then (2) is done. However, here is it where it all goes wrong.

[tex]2 - \frac{p+2}{2^{p}} + \frac{p+1}{2^{p+1}} = 2 - \frac{2(p+2)}{2^{p+1}} + \frac{p+1}{2^{p+1}} = 2 - \frac{2p + 4 + p + 1}{2^{p+1}}[/tex]

Which is not the same as that which is looked for. However, there seem to be a sign error somewhere, since if the last two additions are subtractions, it works. Thank you for your time.