Prove that for all $n\in \mathbb{N}$, $\sum_{i=1}^n i/2^i = 2 - (n+2)/2^n$

Question

Can someone help me with this question please?

Prove that for all $n\in \mathbb{N}$, $$\displaystyle\sum_{i=1}^n \frac{i}{2^i} = 2 - \frac{n+2}{2^n}$$

Answer 1

Consider the following function: $$ f(x)=\sum_{i=1}^n x^i=x\frac{x^{n+1}-1}{x-1} $$ If you differentiate $f$ w.r.t. $x$: $$ f'(x)=\sum_{i=1}^n ix^{i-1}=(x\frac{x^{n+1}-1}{x-1})' $$ Then it is enough to calculate $xf'(x)$ for $x=0.5$.

Answer 2

Using Arithmetico-geometric sequence,

here $a=d=1, r=\frac12$

$$\sum_{i=1}^n \frac i{2^i}=\frac12\sum_{i=1}^n i\cdot \frac1{2^{i-1}} =\frac12\cdot\frac{1-[1+n-1]\frac1{2^n}}{1-\frac12}+\frac12\cdot\frac{\frac12\left(1-\frac1{2^{n-1}}\right)}{\left(1-\frac12\right)^2}$$

$$=1-\frac n{2^n}+1-\frac1{2^{n-1}}=2-\frac n{2^n}-\frac 2{2^n}$$

Answer 3

$$ \sum_{i=1}^n \frac{i}{2^i} = \sum_{i=0}^{n-1} \frac{i+1}{2^{i+1}} =\frac{1}{2}(\sum_{i=0}^{n-1}\frac{i}{2^i} + \sum_{i=0}^{n-1}\frac{1}{2^i}) $$

$$ =\frac{1}{2}(\sum_{i=1}^{n}\frac{i}{2^i} -\frac{n}{2^n}+\frac{1-\frac{1}{2^n}}{\frac{1}{2}}) $$

$$ \frac{1}{2} \sum_{i=1}^n\frac{i}{2^i} = -\frac{n}{2^{n+1}}+1-\frac{1}{2^n} $$ $$ \sum_{i=1}^n\frac{i}{2^i}=-\frac{n}{2^n}+2-\frac{1}{2^{n-1}} $$ $$ =-\frac{n+2}{2^n}+1 $$

Answer 4

Let $u_i = \dfrac{i}{2^i}$ and remark that $u_{i+2}-u_{i+1} + \dfrac{u_i}{4} = 0$ for all $i \geq 1$. Now, it is easy because $$ \frac{1}{4}\sum_{i=1}^n u_i = \sum_{i=1}^n(u_{i+1}-u_{i+2}) = u_2 - u_{n+2}. $$