Given that $\displaystyle Ur = r (2r + 1) + 2^{r + 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 $\displaystyle \frac{1}{6} n(n + 1) (4n + 5) + 2^{(n + 3)} - 4 $

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- Jan 23rd 2008, 07:55 PMgetonConfusion in Series
Given that $\displaystyle Ur = r (2r + 1) + 2^{r + 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 $\displaystyle \frac{1}{6} n(n + 1) (4n + 5) + 2^{(n + 3)} - 4 $ - Jan 23rd 2008, 08:06 PMmr fantastic
$\displaystyle U_r = r(2r + 1) + 2r + 2 = 2r^2 + 3r + 2$.

Therefore $\displaystyle \sum_{r=1}^{n} U_r = \sum_{r=1}^{n} (2r^2 + 3r + 2) = 2 \sum_{r=1}^{n}r^2 + 3 \sum_{r=1}^{n} r + 2n$

Now apply the standard formulae for the sum of first n integers and sum of first n squares of integers ...... - Jan 23rd 2008, 08:54 PMgeton
I’m really very sorry for my mistake.

- Jan 23rd 2008, 09:21 PMbadgerigarQuote:

\frac{1}{6} n(n + 1) (4n + 5) + 2^{(n + 3)} - 4