# Summation Question

1. Nov 10, 2007

### jamesvanboxtel

I have the following recurrence that I am trying to come up with atleast a simplified version if not a closed form.

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

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

If any of you can help that would be awesome.

BTW this forum looks really cool.

Last edited: Nov 10, 2007
2. Nov 11, 2007

### yasiru89

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.

3. Nov 11, 2007

### dodo

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.

4. Nov 11, 2007

### jamesvanboxtel

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