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September 19th, 2007, 05:45 AM   #1
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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.
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September 19th, 2007, 11:03 AM   #2
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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.
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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.
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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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