If f(r) = r(r+1)(r+2), simplify f(r+1)-f(r). Hence find ∑_(r=1)^n▒〖3(r^2+3r+2)〗and deduce ∑_(r=1)^n▒r^2. I stuck at the deduction part, can anyone explain how to do it?
Since f(r) = r(r+1)(r+2), have f(r+1)-f(r) = (r+1)((r+1)+1)((r+1)+2) - r(r+1)(r+2) = (r+1)(r+2)(r+3) - r(r+1)(r+2) = (r+1)(r+2){(r+3)-(r)} = 3(r+1)(r+2).
Thus .
(the last is because the sum telescopes - everything else cancels out.)
Now get f(n+1) - f(1):
Thus
You're presumed to already know that , and .
Now
So pluggin in gives:
Solving for our desired series, , gives:
Thus