Question

Find the sum of the following series upto n terms :

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

Answer 1

The given series, is:

$\frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+\dots \dots \text{upto n terms}$

Let $a_n$ be the nth term of given series and $S_n$ be the sum of the n terms of the given series.

$\therefore a_n=\frac{1^3+2^3+3^3+\dots \dots +n^3}{1+3+5+\dots \dots +(2n-1)}$

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

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

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

$=\frac{1}{4}[1^2+2.1+1]+\frac{1}{4}[2^2+2.2+1]+\dots \dots +\frac{1}{4}[n^2+2n+1]$

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

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

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