Help with a proof.

• Jan 19th 2010, 07:26 PM
seven.j
Help with a proof.
Hi, I'm stuck at proving the following question...

Prove that for all n>0,

1/2 + 2/2^2 + 3/2^3 + ... + n/2^n = 2 - (n+2)/2^n

I've tried all sorts of different ways of solving this, but to no avail.

Any help is appreciated :)
• Jan 19th 2010, 07:50 PM
Drexel28
Quote:

Originally Posted by seven.j
Hi, I'm stuck at proving the following question...

Prove that for all n>0,

1/2 + 2/2^2 + 3/2^3 + ... + n/2^n = 2 - (n+2)/2^n

I've tried all sorts of different ways of solving this, but to no avail.

Any help is appreciated :)

$\sum_{k=1}^{n}\frac{k}{2^k}$. Note that $\sum_{k=1}
^n x^k=\frac{x^{n+1}-x}{x-1}$
. Differentiating both sides and multiplying by $x$ gives $\sum_{k=1}^{n}k\cdot x^k=...$ figure the right side out.
• Jan 19th 2010, 07:54 PM
Krizalid
$\sum\limits_{j=1}^{n}{\frac{j}{2^{j}}}=\sum\limits _{j=1}^{n}{\sum\limits_{k=1}^{j}{\frac{1}{2^{j}}}} =\sum\limits_{k=1}^{n}{\frac{1}{2^{k}}\left( \sum\limits_{j=0}^{n-k}{2^{-j}} \right)}=\frac{1}{2^{n}}\sum\limits_{k=1}^{n}{\fra c{1}{2^{k}}\left( 2^{n+1}-2^{k} \right)},$ you can do the rest, those are finite geometric sums.
• Jan 19th 2010, 08:37 PM
Soroban
Hello, seven!

Here's one way . . .

Quote:

Prove that for all $n>0\!:$

. . $\frac{1}{2}+ \frac{2}{2^2} + \frac{3}{2^3} +\:\hdots\:+ \frac{n}{2^n} \; =\; 2 - \frac{n+2}{2^n}$

$\begin{array}{ccccc}
\text{We have:} &S &=& \dfrac{1}{2} + \dfrac{2}{2^2} + \dfrac{3}{2^3} + \dfrac{4}{2^4} + \:\hdots\:+\dfrac{n}{2^n}\qquad\qquad \\ \\[-3mm]

\text{Multiply by }\dfrac{1}{2}\!: & \dfrac{1}{2}S &=& \quad\;\; \dfrac{1}{2^2} + \dfrac{2}{2^3} + \dfrac{3}{2^4} + \hdots + \dfrac{n-1}{2^n} + \dfrac{n}{2^{n+1}}\end{array}$

. . . $\text{Subtract: }\quad \frac{1}{2}S \;=\;\underbrace{\frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \hdots + \frac{1}{2^n}}_{\text{geometric series}} - \frac{n}{2^{n+2}}$ .[1]

The geometric series has the sum: . $\frac{1}{2}\cdot\frac{1 - \left(\frac{1}{2}\right)^n}{1-\frac{1}{2}} \;=\;1 - \frac{1}{2^n}$

Then [1] becomes: . $\frac{1}{2}S \;=\;\left(1 - \frac{1}{2^n}\right) - \frac{n}{2^{n+1}} \;=\;1 - \frac{n+2}{2^{n+1}}$

Multiply by 2: . $S \;=\;2 - \frac{n+2}{2^n}$

• Jan 19th 2010, 09:19 PM
Jhevon
Quote:

Originally Posted by seven.j
Hi, I'm stuck at proving the following question...

Prove that for all n>0,

1/2 + 2/2^2 + 3/2^3 + ... + n/2^n = 2 - (n+2)/2^n

I've tried all sorts of different ways of solving this, but to no avail.

Any help is appreciated :)

This problem also can be done by induction pretty easily. if you're interested, you can try it that way. to me it was the most knee-jerk approach to try, and it worked out great. but you have lots of nice approaches here to choose from