summation algebra questions

**thereddevils** · February 9th 2009, 01:56 AM

Find the sum of each of the following series .

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

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

**ADARSH** · February 9th 2009, 03:30 AM

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

$<br /> \sum^n_{r=1} {r^3} = \frac{n^2(n+1)^2}{4}<br />$

$<br /> \sum^n_{r=1} {r} =\frac{n(n+1)}{2}<br />$

Use it in your case

**mr fantastic** · February 9th 2009, 03:34 AM

Originally Posted by thereddevils

Find the sum of each of the following series .

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

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

Adding to what ADARSH said:

For (1), note that

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

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

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