Question

Find the sum of the following series to $n$ terms:
$\frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+...$

Answer 1

$S_n=\frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+.....$

$t_n=\frac{1^3+2^3+3^3+.......+n^3}{1+3+5+.....}$

$t_n=\frac{{\left[\frac{n(n+1)}{2}\right]}^2}{n^2}=\frac{n^2{(n+1)}^2}{4n^2}=\frac{1}{4}[n^2+1+2n]=\frac{n^2}{4}+\frac{n}{2}+\frac{1}{4}$

$S_n=\sum t_n$

$=\frac{1}{4}\frac{n(n+1)(2n+1)}{6}+\frac{1}{2}\frac{n(n+1)}{2}+\frac{n}{4}$

$=\frac{n(n+1)}{4}[\frac{2n+1}{6}+1]+\frac{n}{4}$

$=\frac{n(n+1)}{4}\frac{(2n+7)}{6}+\frac{n}{4}$

$=\frac{n}{4}[\frac{(n+1)(2n+7)}{6}+1]$

$=\frac{n}{4}\times \frac{1}{6}[2n^2+7n+2n+7+6]$

$S_n=\frac{n}{24}(2n^2+9n+13)$