Help on series using sigma notation??

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How do I work out : Sigma (2n at the top, and r=n+1 at the bottom) (3r^2-3r+1) = kn^3

k is an integer. 

This is part b of a past paper question. In the previous part I got that Sigma (n at the top, and n=1) (3r^2-3r+1) =n^3. 

Posted Wed 26th September, 2012 @ 18:42 by Natalia

1 Answer

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What you need to find is Sigma (2n at the top, and r=1) for the series and then subtract from it- Sigma (n at the top, and r=1), this should give an answer in the form kn^3, and the resultant series is Sigma (2n at the top, and r=n+1)

Answered Mon 19th November, 2012 @ 17:27 by Nixon
Edited by Nixon on Mon 19th November, 2012 @ 17:27