Thread: summation algebra questions

Find the sum of each of the following series .

(1) $\displaystyle \sum^{2n}_{r=n}r(2r+3)$

(2) $\displaystyle \sum^{3n}_{r=1}r(r^2+1)$

$\displaystyle \sum^n_{r=1} {r^2} = \frac{n(n+1)(2n+1)}{6} $

$\displaystyle
\sum^n_{r=1} {r^3} = \frac{n^2(n+1)^2}{4}
$

$\displaystyle
\sum^n_{r=1} {r} =\frac{n(n+1)}{2}
$

Use it in your case

Originally Posted by thereddevils

Find the sum of each of the following series .

(1) $\displaystyle \sum^{2n}_{r=n}r(2r+3)$

(2) $\displaystyle \sum^{3n}_{r=1}r(r^2+1)$

Adding to what ADARSH said:

For (1), note that

1. $\displaystyle \sum^{2n}_{r=n}r(2r+3) = \sum^{2n}_{r=1}r(2r+3) - \sum^{n-1}_{r=1}r(2r+3) $.

2. $\displaystyle r(2r + 3) = 2r^2 + 3r$ (you should also expand the summand in (2) as well).