Question

Find the sum of the following series to $n$ terms :
$1^3+3^3+5^3+7^3+....$

Answer 1

$\begin{array}{c}{1}^3+5^3+7^3+........+\\ \Rightarrow {\left(2n-1\right)}^3=8n^3-1+6n-12n^2\end{array}$

Let the sum of upto n term of the series

$\begin{array}{c}\Rightarrow S=8\sum n^3-12\sum n^2+6\sum n-\sum 1\\ \Rightarrow 8\frac{n^2{\left(n+1\right)}^2}{4}-12\frac{n\left(n+1\right)\left(2n+1\right)}{6}+6\frac{n\left(n+1\right)}{2}-n\\ \Rightarrow 2n^2{\left(n+1\right)}^2-2n\left(n+1\right)\left(2n+1\right)+3n\left(n+1\right)-n\\ \Rightarrow \left(n+1\right)\left(n\right)\left[2n\left(n+1\right)-2\left(2n+1\right)+3\right]-n\\ \Rightarrow n\left(n+1\right)\left[2x^2+2n-4n-2+3\right]-n\\ \Rightarrow \left(n^2+n\right)\left[2n^2-2n+1\right]-n\\ \Rightarrow 2x^4-2x^3+x^2+2x^3-2x^2+n-n\\ \left(2x^4-x^2\right).\end{array}$