# Thread: closed form is needed

1. ## closed form is needed

I've been searching the Internet for a closed of a following expression :

$\sum_{i=1}^{x} \frac{i}{2^{i}}$

however unsuccessful.

Can anyone shad some light ??

b

pS

even an approximation would do.

sorry for wasting time ,wikipedia has the answer i somehow overlooked it

sorry

2. ## Re: closed form is needed

Originally Posted by baxy77bax
I've been searching the Internet for a closed of a following expression :

$\sum_{i=1}^{x} \frac{i}{2^{i}}$

however unsuccessful.

Can anyone shad some light ??

b

pS

even an approximation would do.
$\displaystyle{\sum_{i=1}^x}\dfrac{i}{2^i}=2^{-x} \left(-x+2^{x+1}-2\right)$

3. ## Re: closed form is needed

Originally Posted by romsek
$\displaystyle{\sum_{i=1}^x}\dfrac{i}{2^i}=2^{-x} \left(-x+2^{x+1}-2\right)$
Just out of interest romsek, how did you evaluate this?

4. ## Re: closed form is needed

Originally Posted by Prove It
Just out of interest romsek, how did you evaluate this?
I cheated and dumped it into Mathematica

5. ## Re: closed form is needed

Hello, baxy77bax!

Find a closed form for: . $\sum^x_{i=1}\frac{i}{2^i}$

$\begin{array}{ccccccc}\text{We have:} & S &=& \frac{1}{2} + \frac{2}{2^2} + \frac{3}{2^3} + \frac{4}{2^4} + \frac{5}{2^5} + \cdots + \frac{x}{2^x}\qquad\quad & [1] \\ \text{Multiply by }\frac{1}{2}\!: & \frac{1}{2}S &=& \quad\quad \frac{1}{2^2} + \frac{2}{2^3} + \frac{3}{2^4} + \frac{4}{2^5} + \cdots + \frac{x-1}{2^x} + \frac{x}{2^{x+1}} & [2] \\ \text{Subtract [1]-[2]:} & \frac{1}{2}S &=& \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \frac{1}{2^5} + \cdots + \frac{1}{2^x} - \frac{x}{2^{x+1}} \end{array}$

Hence: . $\frac{1}{2}S \;=\;\left(1 - \frac{1}{2^x}\right) - \frac{x}{2^{x+1}} \quad\Rightarrow\quad S \;=\;2 - \frac{1}{2^{x-1}} - \frac{x}{2^x}$

Therefore: . $S \;=\;\frac{1}{2^x}\left(2^{x+1} - 2 - x\right)$