Question

Find the sum of series up 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

The $n^{th}$ term of the given series is

$\frac{1^3+2^3+3^3+...+n^3}{1+3+5+...+(2n-1)}=\frac{{\left[\frac{n(n+1)}{2}\right]}^2}{1+3+5+...+(2n-1)}$

Here $,1,3,5,...(2n-1)$ is an A.P. with first term $a$ , last term $(2n-1)$ and number of terms as $n$

$\therefore 1+3+5+...+(2n-1)=\frac{n}{2}[2\times 1+(n-1\left)2\right]=n^2$

$\therefore a_n=\frac{n^2{(n+1)}^2}{{4n}^2}=\frac{{(n+1)}^2}{4}=\frac{1}{4}{n}^2+\frac{1}{2}n+\frac{1}{4}$

$\therefore S_n=\sum _{k=1}^n{a}_k=\sum _{k=1}^n(\frac{1}{4}{K}^2+\frac{1}{2}K+\frac{1}{4})$

By using ${\sum }_{i=1}^{n}k=1+2+3+\dots +n=\frac{n(n+1)}{2}$ and

${\sum }_{i=1}^n{k}^2=1^2+2^2+3^2+\dots +n^2=\frac{n(n+1)(2n+1)}{6}$ we have,

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

$=\frac{n\left[\right(n+1\left)\right(2n+1)+6(n+1\left)6\right]}{24}$

$=\frac{n[{2n}^2+3n+1+6n+6+6]}{24}$

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