Question

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

Answer 1

THe General term for this expression is$=\frac{{\sum }_{k=1}^{k=r}{k}^3}{r}$

We know that $\sum _{k=1}^{k=r}{k}^3=\frac{r^2(r+1{)}^2}{4}$

Now solving the General term again we get

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

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

General term$=\frac{r^3+2r^2+r}{4}$

Sum of the given series is $\sum _{r=1}^{r=n}\frac{r^3+2r^2+r}{4}=\frac{1}{4}(\sum _{r=1}^{r=n}{r}^3+2\sum _{r=1}^{r=n}{r}^2+\sum _{r=1}^{r=n}r)$

Using the formula

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

$\sum _{r=1}^{r=n}{r}^2=\frac{n(n+1)(2n+1)}{6}$

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

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