1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Summation Question

  1. Nov 10, 2007 #1
    I have the following recurrence that I am trying to come up with atleast a simplified version if not a closed form.

    [tex]T(n) = T(n-1) + \sum_{i=1}^{(n-1)/2} [(n-(i+1)) * (i-1) * 2 + 2][/tex]

    in addition if n is even I must add the following to T(n)
    [tex]((n/2) - 1)^2[/tex]

    If any of you can help that would be awesome.

    BTW this forum looks really cool.
    Last edited: Nov 10, 2007
  2. jcsd
  3. Nov 11, 2007 #2
    This isn't all that clear but why not move the term on the right and have T(n) - T(n-1)
    then sum over n from 1 to n (if summing over n with top limit n is confusing index to a different letter, it won't make a difference in the end)on both sides for a double sum on the right.
    Then the left will be have T(n) - T(0) , the T(0) you may know.
    As for the right since only multiplication seems to be involved prominently the original summational need not be a problem though I'm not sure of the second that I suggested(over n)
    You'll need to consider seperate cases for n odd or even also. Hope I interpreted correctly.
  4. Nov 11, 2007 #3
    Of course, the constant +2 can be taken out of the sum; summing +2 in a loop, 'm' times, is just adding 2*m.

    One question: do you really need each and all T(n)? Because maybe you could sum the terms in pairs, that is, construct a new series 'A' where A(0)=T(0)+T(1), A(1)=T(2)+T(3), A(2)=T(4)+T(5) and so on, so that the special case for even/odd terms dissapears.
  5. Nov 11, 2007 #4
    Turns out I was able to enter the first few results here and come up with what the pattern was.

    http://www.research.att.com/~njas/sequences/ [Broken]

    Thanks for trying though guys.
    Last edited by a moderator: May 3, 2017
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook