# Thread: Question about summation

1. ## Question about summation

Hi, i'm not sure on how to do the summation of the following :

$\displaystyle n$
$\displaystyle \sum (1-\frac{i-1}{n}).\frac{1}{n}$
$\displaystyle i=1$

$\displaystyle n$
$\displaystyle \sum (1-\frac{i}{n}).\frac{1}{n}$
$\displaystyle i=1$

The first one i got the answer is 1/2 (1 - 1/n) after simplifying everything.
I'm stuck with the second one , not sure on how to sum up everything.

2. $\displaystyle \sum_{i\,=\,1}^n\left(1-\frac{i-1}n\right)\frac1n$

$\displaystyle =\ \frac1n\sum_{i\,=\,1}^n1-\frac1{n^2}\sum_{i\,=\,1}^n(i-1)$

$\displaystyle =\ \frac1n(n)-\frac1{n^2}\cdot\frac{n(n-1)}2$

$\displaystyle =\ 1-\frac{n-1}{2n}$

$\displaystyle =\ \frac{n+1}{2n}$

The second one is

$\displaystyle \sum_{i\,=\,1}^n\left(1-\frac in\right)\frac1n$

$\displaystyle =\ \sum_{i\,=\,2}^{n+1}\left(1-\frac{i-1}n\right)\frac1n$

$\displaystyle =\ \sum_{i\,=\,1}^{n}\left(1-\frac{i-1}n\right)\frac1n\,+\left(1-\frac {(n+1)-1}n\right)\frac1n-\left(1-\frac{1-1}n\right)\frac1n$

$\displaystyle =\ \frac{n+1}{2n}+0-\frac1n$

$\displaystyle =\ \frac{n-1}{2n}$

3. Originally Posted by xcluded
Hi, i'm not sure on how to do the summation of the following :

$\displaystyle n$
$\displaystyle \sum (1-\frac{i-1}{n}).\frac{1}{n}$
$\displaystyle i=1$

$\displaystyle n$
$\displaystyle \sum (1-\frac{i}{n}).\frac{1}{n}$
$\displaystyle i=1$

The first one i got the answer is 1/2 (1 - 1/n) after simplifying everything.
I'm stuck with the second one , not sure on how to sum up everything.

$\displaystyle \sum_{i=1}^n\left(1-\frac{i}{n}\right)\frac{1}{n}=\frac{1}{n}\left[ \sum_{i=1}^n\,1-\frac{1}{n}\sum_{i=1}^n\,i\right]=\frac{1}{n}\left[n-\frac{1}{n}\frac{n(n+1)}{2}\right]=....$

Tonio

4. Thanks guys !