Question

$\frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+....n$ terms $=\frac{n}{24}(2n^2+9n+13)$.

Answer 1

We have,

$\frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+........$

We can write $n^{th}$ term of series is

$a_n=\frac{1^3+2^3+3^3+.....+n^3}{1+3+5+......+nterms}$

Also, we know that,

$1^3+2^3+3^3+......+n^3$

$={\sum }_{n=1}^n{n}^3$

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

Again,

$1+3+5+.....+n$

This is an $A.P.$

whose first term is $a=1$

common difference $\left(d\right)=3-1=2$

Now,

Sum of series$=S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

$=\frac{n}{2}[2\times 1+(n-1)\times 2]$

$S_n=n^2$

Now,

$a_n=\frac{1^3+2^3+3^3+.....+n^3}{1+3+5+.......+nterms}$

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

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

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

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

$=\frac{n^2+2n+1}{4}$

Now, finding sum of series

$S_n={\sum }_{n=1}^n{a}_n$

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

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

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

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

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

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

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

$\frac{n}{24}[2n^2+3n+1+6n+12]$

$=\frac{n}{24}[2n^2+9n+13]$

Hence, proved.