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:

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

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- Jan 31st 2009, 08:52 AMTwenty80Summation 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 AMred_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 PMTwenty80
- Jan 31st 2009, 12:31 PMzhupolongjoe
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 PMTwenty80
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 PMzhupolongjoe
Check this out, hopefully you can apply the same logic to your problem:

Summations - Jan 31st 2009, 05:20 PMTwenty80
- Jan 31st 2009, 05:53 PMmr fantastic
- Jan 31st 2009, 06:25 PMzhupolongjoe
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 PMTwenty80
haha, that's sad. You guys were right, I didn't realize it was the same thing but backwards.