Question

Sum to $n$ terms the series :
$1\times 3+3\times 5+5\times 7+7\times 9+..$

Answer 1

Let the $n^{th}$ term be denoted by $t_n$

$\therefore t_n=\left(\{1,3,5,...\}\right)\times \left(\{3,5,7,...\}\right)$

$t_n$ of $1,3,5,7,...$ is $2n-1$

and $t_n$ of $3,5,7,..$ is $2n+1$

$\therefore t_n$ of the above series is $(2n-1)(2n+1)=4n^2-1$

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

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

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

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

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

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

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