• October 22nd 2009, 08:57 AM
xcluded
Hi, i'm not sure on how to do the summation of the following :

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

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

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.
• October 22nd 2009, 09:16 AM
proscientia
$\sum_{i\,=\,1}^n\left(1-\frac{i-1}n\right)\frac1n$

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

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

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

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

The second one is

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

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

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

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

$=\ \frac{n-1}{2n}$
• October 22nd 2009, 09:17 AM
tonio
Quote:

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

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

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

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.

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

Tonio
• October 22nd 2009, 09:33 AM
xcluded
Thanks guys !