# double summation

• Mar 30th 2009, 10:40 AM
premierplayer
double summation
Been trying to get this problem for a few hours. Been going through the book and online for some examples but haven't found anything that makes sense.

If someone could help me that would be gooood.

Compute the double sum:

http://img407.imageshack.us/img407/5117/summation.jpg
• Mar 30th 2009, 12:02 PM
Soroban
Hello, premierplayer!

If you don't understand a basic summation,

Quote:

Compute the double sum: . $\sum^3_{i=1} \sum^2_{j=1} (i-j)$

In the inner summation, let $j = 1, 2$ and add them up.

. . $\sum^2_{j=1}(i-j) \;=\;(i-1) + (i-2) \;=\;2i - 3$

Now that goes in the outer summation: . $\sum^3_{i=1}(2i - 3)$
Let $i = 1,2,3$ and add them up.

. $\sum^3_{i=1}(2i-3) \;=\;\bigg[2(1)-3\bigg] + \bigg[2(2)-3\bigg] + \bigg[2(3)-3\bigg] \;=\;-1 + 1 + 3 \;=\;3$

• Mar 30th 2009, 12:11 PM
premierplayer
Quote:

Originally Posted by Soroban
Hello, premierplayer!

If you don't understand a basic summation,

In the inner summation, let $j = 1, 2$ and add them up.

. . $\sum^2_{j=1}(i-j) \;=\;(i-1) + (i-2) \;=\;2i - 3$

Now that goes in the outer summation: . $\sum^3_{i=1}(2i - 3)$
Let $i = 1,2,3$ and add them up.

. $\sum^3_{i=1}(2i-3) \;=\;\bigg[2(1)-3\bigg] + \bigg[2(2)-3\bigg] + \bigg[2(3)-3\bigg] \;=\;-1 + 1 + 3 \;=\;3$

Awesome.

So I pressume that the number above the sumation has to do with how many times we calculate.

for example 2 is (i-1) + (i-2)... etc..

I get it. Thanks.