Question

Find the sum of $n$ terms of the series whose $nth$ term is :
$(2n-1{)}^2$

Answer 1

$\sum (2n-1{)}^2$

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

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

$\frac{4n\left(n+1\right)(2n+1)}{6}+\frac{4n(n+1)}{2}+n$