Question

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

Answer 1

$k$th term of the series is $=(2k-1)(2k+1)$ $=4k^2-1$

Sum of first $n$ terms is,

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

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

$=\frac{4n^3+6n^2-n}{3}$

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