Given that , find n ∑Ur r=1 I’ve done but I’m confused about my answer. Please tell me is it wrong or right. I got
Last edited by geton; Jan 23rd 2008 at 08:37 PM. Reason: Correcting 2r + 2
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Originally Posted by geton Given that Ur = r (2r + 1) + 2r + 2 , find n ∑Ur r=1 I’ve done but I’m confused about my answer. Please tell me is it wrong or right. I got . Therefore Now apply the standard formulae for the sum of first n integers and sum of first n squares of integers ......
I’m really very sorry for my mistake.
\frac{1}{6} n(n + 1) (4n + 5) + 2^{(n + 3)} - 4 Looks right to me.
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