Question

1.3+3.5+5.7+....+$\left(2n-1\right)\left(2n+1\right)=\frac{n\left(4n^2+6n-1\right)}{3}$

Answer 1

$S=1.3+3.5+....+(2n-1)(2n+1)S={\sum }_{k=1}^n(2k-1)(2k+1)={\sum }_{k=1}^n({4k}^2-1)=4{\sum }_{k=1}^n{k}^2-{\sum }_{k=1}^{n}1S\frac{4n(n+1)(2n+1)}{6}-n=n(\frac{2({2n}^2+3n+1)}{3}-1)=n\left(\frac{({4n}^2+6n-1)}{3}\right)S=n\left(\frac{({4n}^2+6n-1)}{3}\right)$