Results 1 to 10 of 10

Math Help - Summation Help

  1. #1
    Newbie
    Joined
    Jan 2009
    Posts
    8

    Summation Help

    I'm a little confused by summations and closed forms...

    Can someone explain how the heck this works...
    \sum_{i=1}^{n}n-i+2


    By re-arranging the terms, this becomes:

    \sum_{i=1}^{n}i+1
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor red_dog's Avatar
    Joined
    Jun 2007
    From
    Medgidia, Romania
    Posts
    1,245
    Thanks
    1
    \sum_{i=1}^n(n-i+2)=(n+1)+n+(n-1)+\ldots+2=

    =2+3+\ldots+n+(n+1)=\sum_{i=1}^n(i+1)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2009
    Posts
    8
    Quote Originally Posted by red_dog View Post
    \sum_{i=1}^n(n-i+2)=(n+1)+n+(n-1)+\ldots+2=

    =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.
    Last edited by mr fantastic; December 8th 2009 at 11:57 AM. Reason: rt --> de
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Jan 2009
    Posts
    296
    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.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jan 2009
    Posts
    8
    I hate to say this, but I'm still not following. It would be great if someone could step through it for me..
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Jan 2009
    Posts
    296
    Check this out, hopefully you can apply the same logic to your problem:

    Summations
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Jan 2009
    Posts
    8
    Quote Originally Posted by zhupolongjoe View Post
    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)
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Twenty80 View Post
    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 View Post
    \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 View Post
    =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.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Senior Member
    Joined
    Jan 2009
    Posts
    296
    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?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Jan 2009
    Posts
    8
    haha, that's sad. You guys were right, I didn't realize it was the same thing but backwards.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. How to do a Summation
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: May 14th 2011, 11:08 AM
  2. Summation
    Posted in the Pre-Calculus Forum
    Replies: 6
    Last Post: May 7th 2011, 06:23 AM
  3. summation help
    Posted in the Calculus Forum
    Replies: 3
    Last Post: February 16th 2009, 02:08 PM
  4. summation
    Posted in the Algebra Forum
    Replies: 1
    Last Post: February 2nd 2009, 08:17 AM
  5. summation
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: October 8th 2008, 04:19 PM

Search Tags


/mathhelpforum @mathhelpforum