Question

Find the sum $\frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+\dots$ to n terms

Answer 1

$\frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+\dots tonterms$
$T_r=\frac{1^3+2^3+3^3\dots r^3}{1+3+5+(2r-1)}=\frac{{\left[\frac{r(r+1)}{2}\right]}^2}{r^2}$
$\therefore T_4=\frac{(r+1{)}^2}{4}=\frac{r^2+2r+1}{4}$
$S_n=\sum T_r=\frac{1}{4}\sum (r^2+2r+1)$
$\therefore S_n=\frac{1}{4}[\sum n^2+2\sum n+\sum 1]$
$\therefore S_n=\frac{1}{4}[\frac{n(n+1)(2n+1)}{6}+2\frac{n(n+1)}{2}+n]$
$\therefore S_n=\frac{1}{4}[\frac{n(2n+1)(n+1)}{6}+n^2+n+n]$
$=\frac{1}{4}\left[\frac{n(n+1)(2n+1)+6n^2+12n}{6}\right]$
$=\frac{1}{4}\left[\frac{(n^2+n)(2n+1)+6n^2+12n}{6}\right]$
$=\frac{1}{4}\left[\frac{2n^3+n^2+2n^2+n+6n^2+12n}{6}\right]$
$=\frac{1}{4}\left[\frac{2n^3+9n^2+13n}{6}\right]$
$\therefore S_n=\frac{n(2n^2+9n+13)}{24}$