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