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 September 19th, 2007, 05:45 AM #1 Newbie   Joined: Sep 2007 Posts: 17 Thanks: 0 help with the calculus of infinite sequence I need to know step-by-step solution how to get the formula (on the right) from this sum of infinite sequence (on the right). can anyone help me? thank you all very much in advance.
 September 19th, 2007, 11:03 AM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms I don't think you mean infinite -- the sum only runs to n. If it was infinite you'd be looking at zeta(-4) which is a trivial zero by analytic continuation. You can solve this by induction. Show that it works for n = 1, and that if it holds for n it also holds for n+1. Do this by adding n(n+1)(2n+1)(3n^2+3n-1)/30+(n+1)^4 and simplifying.
September 19th, 2007, 10:44 PM   #3
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thank you very much! yes, youre right this is not an infinite seqeunce, im sorry, but can you give me just one more advise how you came to this?

Quote:
 Originally Posted by CRGreathouse n(n+1)(2n+1)(3n^2+3n-1)/30+(n+1)^4
thanks.

September 20th, 2007, 08:06 AM   #4
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Quote:
 Originally Posted by 1210 but can you give me just one more advise how you came to this?
Expand the formula I gave you, then re-collect terms so the final formula is (n+1)(n+2)(2(n+1)+1)(3(n+1)^2+3(n+1)-1)/30.

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