Question

Find the sum of the following series to $n$ terms: $\frac{1^3}{1}+\frac{1^3+2^3}{1+3}+\frac{1^3+2^3+3^3}{1+3+5}+\dots$

Answer 1

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

$t_n=\frac{1^3+2^3+3^3+...+n^3}{1+3+5+...+(2n-1)}$

$1^3+2^3+3^3+...+n^3=\frac{n^2{(n+1)}^2}{4}$

$1+3+5+...+(2n-1)=n^2$

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

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

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

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

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

$=\frac{n(n+1)(2n+7)+6n}{24}$

$=\frac{n\left[\right(n+1\left)\right(2n+7)+6]}{24}$

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