Finding sum of none arithmetic series

Each term in the equation is $n(2n-1)$, so $$S=\sum_{n=1}^{100}{n(2n-1)}=2\sum_{n=1}^{100} n^2-\sum_{n=1}^{100} n=2\frac{m(m+1)(2m+1)}{6}-\frac{m(m+1)}{2}$$ where $m=100$.

By the telescoping sum we obtain: $$\sum_{n=1}^{100}n(2n-1)=\sum_{n=1}^{100}\left(\frac{n(n+1)(4n-1)}{6}-\frac{(n-1)n(4n-5)}{6}\right)=$$ $$=\frac{100\cdot101\cdot399}{6}=671650.$$

$a_n=\sum_{r=1}^n(4r-3)+a_0=\dfrac n2(1+4n-3)+a_0=2n^2-n+a_0$

$$\sum_{n=1}^ma_n=2\sum_{n=1}^mn^2-\sum_{n=1}^mn+a_0\sum_{n=1}^m1$$

Here $a_0=0$

Alternatively,

$$a_m=b_m+a+bm+cm^2$$

$$4n-3=a_n-a_{n-1}=b_n-b_{n-1}+b+c(2n-1)=b_n-b_{n-1}+2c(n)+b-c$$

WLOG set $2c=4,b-c=-3\iff c=b+3$ to find $b_n=b_{n-1}$

set $a=0$ so that $b_m=a_m=0$