Question

Find the sum of $1^3+3^3+5^3+7^3+\cdots$ to $n$ terms.

Answer 1

$S=1^3+3^3+5^3+...+{(2n-1)}^3=1^3+2^3+3^3+4^3+5^3+...+{(2n-1)}^3+{\left(2n\right)}^3-(2^3+4^3+6^3+...+{\left(2n\right)}^3)$

Sum of cube of $n$ natural numbe will be ${\left(\frac{n(n+1)}{2}\right)}^2$

Sum of cube of $2n$ natural numbe will be ${\left(\frac{2n(2n+1)}{2}\right)}^2$

$S=1^3+3^3+5^3+...+{(2n-1)}^3={\left(\frac{2n(2n+1)}{2}\right)}^2-2^3(1^3+2^3+3^3+4^3+5^3+...+n^3)$

$S=1^3+3^3+5^3+...+{(2n-1)}^3=4\frac{n^2}{4}({(2n+1)}^2-2{(n+1)}^2)$

$S=1^3+3^3+5^3+...+{(2n-1)}^3=n^2(4n^2+1+4n-2n^2-4n-2)$

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