Sum Sum

Level 2

∑ n = 1 ∞ n 2 n = ? \large \sum_{n=1}^{\infty}\frac{n}{2^n}=?

4 2 1 1 4 \frac 14 1 2 \frac 12

Ar Hakim by AR Hakim , 25, Indonesia

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Chew-Seong Cheong
Oct 29, 2018

S = ∑ n = 1 ∞ n 2 n = ∑ n = 0 ∞ n 2 n = ∑ n = 0 ∞ n + 1 2 n + 1 = 1 2 ∑ n = 0 ∞ n + 1 2 n = 1 2 ∑ n = 0 ∞ n 2 n + 1 2 ∑ n = 0 ∞ 1 2 n = 1 2 S + 1 2 ( 1 1 − 1 2 ) = 1 2 S + 1 = 2 \begin{aligned} S & = \sum_{\color{#3D99F6}n=1}^\infty \frac n{2^n} = \sum_{\color{#D61F06}n=0}^\infty \frac n{2^n} = \sum_{\color{#D61F06}n=0}^\infty \frac {n+1}{2^{n+1}} = \frac 12 \sum_{n=0}^\infty \frac {n+1}{2^n} \\ & = \frac 12 \sum_{n=0}^\infty \frac n{2^n} + \frac 12 \sum_{n=0}^\infty \frac 1{2^n} \\ & = \frac 12S + \frac 12 \left(\frac 1{1-\frac 12}\right) \\ & = \frac 12S + 1 \\ & = \boxed 2 \end{aligned}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...