# Summation Help

• Jan 31st 2009, 08:52 AM
Twenty80
Summation Help
I'm a little confused by summations and closed forms...

Can someone explain how the heck this works...
Quote:

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

By re-arranging the terms, this becomes:

$\displaystyle \sum_{i=1}^{n}i+1$
• Jan 31st 2009, 11:37 AM
red_dog
$\displaystyle \sum_{i=1}^n(n-i+2)=(n+1)+n+(n-1)+\ldots+2=$

$\displaystyle =2+3+\ldots+n+(n+1)=\sum_{i=1}^n(i+1)$
• Jan 31st 2009, 12:15 PM
Twenty80
Quote:

Originally Posted by red_dog
$\displaystyle \sum_{i=1}^n(n-i+2)=(n+1)+n+(n-1)+\ldots+2=$

$\displaystyle =2+3+\ldots+n+(n+1)=\sum_{i=1}^n(i+1)$

Maybe I'm just having a brain fade, but how did you get from (n+1)+n+(n-1).... to 2+3+...n+(n+1) to the final given summation.
• Jan 31st 2009, 12:31 PM
zhupolongjoe
You just have to look at it and see.

The first term of the new summation is i=1, i+1=1+1=2

i=2, i+1=2+1=3
.....
i=n, i+1=n+1

Pretty much you just have to write out the terms and see how it goes.
• Jan 31st 2009, 04:04 PM
Twenty80
I hate to say this, but I'm still not following. It would be great if someone could step through it for me..
• Jan 31st 2009, 05:01 PM
zhupolongjoe
Check this out, hopefully you can apply the same logic to your problem:

Summations
• Jan 31st 2009, 05:20 PM
Twenty80
Quote:

Originally Posted by zhupolongjoe
Check this out, hopefully you can apply the same logic to your problem:

Summations

Thanks for the link but I already understand the basics of Summation. I'm sure this may come as a shock to some of you..

My confusion is how does (n+1) become 2, and then n becomes 3...

How does n + (n -1) become n + (n + 1)
• Jan 31st 2009, 05:53 PM
mr fantastic
Quote:

Originally Posted by Twenty80
Thanks for the link but I already understand the basics of Summation. I'm sure this may come as a shock to some of you..

My confusion is how does (n+1) become 2, and then n becomes 3...

How does n + (n -1) become n + (n + 1)

Quote:

Originally Posted by red_dog
$\displaystyle \sum_{i=1}^n (n-i+2) = (n+1)+n+(n-1)+\ldots+2=$

I assume you can see where each term comes from. Substitute i = 1, 2, 3, .....

Quote:

Originally Posted by red_dog
$\displaystyle =2+3+\ldots+n+(n+1)=\sum_{i=1}^n(i+1)$

This is just the above written backwards. The summand on the right hand side should be obvious.
• Jan 31st 2009, 06:25 PM
zhupolongjoe
I see what confuses you. (n+1) does not "become" 2 and n does not
"become" 3. 2+3+....(n-2)+(n-1)+(n)+(n+1) is exactly the same as (n+1)+(n)+(n-1)+(n-2)+....+3+2...just backwards. Do you see it now?
• Jan 31st 2009, 07:47 PM
Twenty80
haha, that's sad. You guys were right, I didn't realize it was the same thing but backwards.