Question

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

Answer 1

Given,

${\sum }_{r=1}^n\frac{1^3+2^3+.....+r^3}{r+1}$

Formula, $1^3+2^3+.....+r^3={\left(\frac{r(r+1)}{2}\right)}^2$

$\therefore {\sum }_{r=1}^n\frac{1^3+2^3+.....+r^3}{r+1}={\left(\frac{r}{2}\right)}^2(r+1)$

$=\frac{1}{4}\cdot {\sum }_{r=1}^n{r}^2(r+1)$

$=\frac{1}{4}\cdot {\sum }_{r=1}^n{r}^3+r^2$

$=\frac{1}{4}\cdot ({\sum }_{r=1}^n{r}^3+{\sum }_{r=1}^n{r}^2)$

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

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

$=\frac{1}{4}\cdot \frac{3n^4+10n^3+9n^2+2n}{12}$

$=\frac{3n^4+10n^3+9n^2+2n}{48}$