+118609 Hi Old Timer 
Sum of n terms for -3n^2+2n+5
\(\displaystyle\sum_{m=1}^n\; (-3m^2+2m+5)\\ =\displaystyle\sum_{m=1}^n\; -3m^2\quad+\displaystyle\sum_{m=1}^n2m\quad +5n\\ =\displaystyle\sum_{m=1}^n\; -3m^2\quad+\displaystyle\sum_{m=1}^n2m\quad +5n\\ \qquad\qquad \displaystyle\sum_{m=1}^n2m=\frac{n}{2}(a+L) = \frac{n}{2}(2+2n) = n(1+n)=n^2+n\\ =-3\left[\displaystyle\sum_{m=1}^n\; m^2\right]\quad+n^2+n\quad +5n\\ \text{** the next line was taken from a khan academy video which I will reference at the end.}\\ =-3\left[\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}\right]\quad+n^2+n\quad +5n\\ =\left[-n^3-\frac{3n^2}{2}-\frac{n}{2}\right]\quad+n^2+n\quad +5n\\ =\frac{-2n^3}{2}-\frac{3n^2}{2}-\frac{n}{2}\quad+\frac{2n^2+2n\quad +10n}{2}\\~\\ =\dfrac{ -2n^3-n^2 +11n }{2} \)
Khan Academy videos:
https://www.khanacademy.org/math/calculus-home/series-calc/series-basics-challenge/v/sum-n-squares-2
You will need to check the algebra, I could easily have made a careless mistake :)
