
summation
Find expression in terms of n for the sum of this series , to the number of terms stated , simplifynig the answer as far as possible .
$\displaystyle 2^3+4^3+6^3+8^3+...$ to n terms .
My work :
$\displaystyle \sum^{n}_{r=1}r^3\sum^{n}_{r=1} (2r1)^3$
$\displaystyle =\frac{1}{4}n^2(n+1)^2\sum^{n}_{r=1}(8r^312r^2+6r1)$
$\displaystyle =\frac{1}{4}n^2(n+1)^28[\frac{1}{4}n^2(n+1)^2]+12(\frac{1}{6}n(n+1)(2n+1))6(\frac{1}{2}n(n+1))+n$
I am not really sure how to put it into its simplest form from here .
i can only see that n is common .

$\displaystyle (2\cdot 1)^3+(2\cdot 2)^3+(2\cdot 3)^3+\ldots+(2n)^3=$
$\displaystyle =\sum_{k=1}^n(2k)^3=8\sum_{k=1}^nk^3=8\cdot\frac{n ^2(n+1)^2}{4}=2n^2(n+1)^2$