Question

Find the sum of $n$ terms of the series $1.3+3.5+5.7+...$

Answer 1

In the given series every term has two factor.If we write all the first factors of each terms in the same order $1,3,5,7,...$ form an A.P

The $n^{th}$ term of this A.P$=1+(n-1)2=2n-1$

The $n^{th}$ term of $3,5,7,...$ is $=3+(n-1)2=2n+1$

$\therefore$ the $n^{th}$ of the given series$=\sum (2n-1)(2n+1)$

$=\sum (4n^2-1)$

$=4\sum n^2-\sum 1$

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

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

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

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

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

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

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