Finding sum of none arithmetic series I have a question to find the sum of the following sum:
$$
S = \small{1*1+2*3+3*5+4*7+...+100*199}
$$
I figured out that for each element in this series the following holds:
$$
a_n = a_{n-1} + 4n - 3
$$
But I don't know where to go from here, I tried subtracting some other series but that did not work very well

$a_n=\sum_{r=1}^n(4r-3)+a_0=\dfrac n2(1+4n-3)+a_0=2n^2-n+a_0$
$$\sum_{n=1}^ma_n=2\sum_{n=1}^mn^2-\sum_{n=1}^mn+a_0\sum_{n=1}^m1$$
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 $b_m=a_m=0$
