Question

Solve
$1^3+\frac{1^3+2^3}{2}+\frac{1^3+2^3+3^3}{3}+...$

Answer 1

We have,

$1^3+\frac{1^3+2^3}{2}+\frac{1^3+2^3+3^3}{3}+........$

let

$T_r=\frac{1^3+2^3+3^3+4^3+........}{1+2+3+\mathrm{4.......}}$

$=\frac{{\left(\frac{r(r+1)}{2}\right)}^2}{r}$

$S_n=\sum T_r=\frac{1}{4}r(r+1{)}^2$

$=\frac{1}{4}r(r^2+1^2+2r)$

$=\frac{1}{4}[r^3+r+2r^2]$

$=\frac{1}{4}[\sum r^3+\sum r+2\sum r^2]$

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

$=\frac{1}{4}\left[\frac{3n^2(n+1{)}^2+4n(n+1\left)\right(2n+1)+6n(n+1)}{12}\right]$

$=\frac{1}{4}\left[\frac{3n^2(n^2+1^2+2n)+4n[2n^2+n+2n+1]+6n^2+6n}{12}\right]$

$=\frac{1}{4}\left[\frac{3n^4+3n^2+6n^3+8n^3+4n^2+8n^2+4n+6n^2+6n}{12}\right]$

$=\frac{1}{48}[3n^4+14n^3+21n^2+10n]$

Hence, this is the answer.