Find a formula for
Sum_{j=1}^n (j+3) (j-4)
So far, I have foiled to j^2-j-12. Thus,Sum_{j=1}^n(j^2) - Sum_{j=1}^n(j) - Sum_{j=1}^n(12).
From Special Sum formulas, I have gotten to
n(n+1)(2n+1)/6 - n(n+1)/2 - 12n
What do I do from here?
Huh? You have the answer. Just simplify if you want:
n(n^2-37)/3