Results 1 to 2 of 2

Math Help - Summation Question

  1. #1
    Newbie
    Joined
    Feb 2009
    Posts
    1

    Summation Question

    \sum^{n-3}_{j=0}(2(n+j-2))

    = 2 * ( (n-2) + (n-1) + n +... + (2n-5) )

    = 2 * ( \sum^{2n-5}_{i=1}(i) - \sum^{n-3}_{i=1}(i) )
    ...

    I don't understand where the (2n-5) comes from in line 2, and then why we use it in the summation on line 3. Nor do I understand where those summations come from on line 3 and why we subtract them.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Mar 2007
    Posts
    1,240

    Talking

    When j = n - 3, what is the simplified form of n + j - 5?

    If you can simplify "n + j - 2" to be something simpler, why not?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Summation question
    Posted in the Algebra Forum
    Replies: 5
    Last Post: February 2nd 2011, 06:26 AM
  2. summation question
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: October 2nd 2010, 06:53 AM
  3. summation question
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: June 21st 2010, 04:57 PM
  4. Summation Question
    Posted in the Algebra Forum
    Replies: 4
    Last Post: May 26th 2010, 03:15 PM
  5. summation question
    Posted in the Algebra Forum
    Replies: 4
    Last Post: May 25th 2007, 10:38 AM

Search Tags


/mathhelpforum @mathhelpforum