Given $t_n=(2n-1)^2=4n^2-4n+1$
We know that sum of $n$ terms of the series $=S_n=\sum t_n=\sum(4n^2-4n+1)$
$\Rightarrow\:S_n=4\sum n^2-4.\sum n+\sum 1$
$\qquad\:=4.\large\frac{n(n+1)(2n+1)}{6}$$-4.\large\frac{n(n+1)}{2}$$+n$
$\qquad\:=n\bigg[\large\frac{2(n+1)(2n+1)}{3}$$-2(n+1)+1\bigg]$
$\qquad\:=n\bigg[\large\frac{4n^2+6n+2-6n-6+3}{3}\bigg]$
$\qquad\:=\large\frac{n(4n^2-1)}{3}$