Confusion in Series

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• January 23rd 2008, 07:55 PM
geton
Confusion in Series
Given that $Ur = r (2r + 1) + 2^{r + 2}$, 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 $\frac{1}{6} n(n + 1) (4n + 5) + 2^{(n + 3)} - 4$
• January 23rd 2008, 08:06 PM
mr fantastic
Quote:

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

$U_r = r(2r + 1) + 2r + 2 = 2r^2 + 3r + 2$.

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

Now apply the standard formulae for the sum of first n integers and sum of first n squares of integers ......
• January 23rd 2008, 08:54 PM
geton
I’m really very sorry for my mistake.
• January 23rd 2008, 09:21 PM
badgerigar
Quote:

\frac{1}{6} n(n + 1) (4n + 5) + 2^{(n + 3)} - 4
Looks right to me.