Question

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

Answer 1

Given $1.3+3.5+5.7+....+(2n-1)(2n+1)$

$\Rightarrow {\sum }_{n=1}^n(2n-1)(2n+1)$

$\Rightarrow {\sum }_{n=1}^{n}4n^2-1$

$\Rightarrow 4{\sum }_{n=1}^w{n}^2-{\sum }_{n=1}^{n}1$

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

$\Rightarrow n[\frac{2(n+1)(2n+1)}{3}-1]$

$\Rightarrow n\left[\frac{2(2n^2+n+2n+1)-3}{3}\right]$

$\Rightarrow \frac{n(4n^2+6n-1)}{3}$

Hence proved