# Strong Induction - Integrals

1. Jan 28, 2009

### XJellieBX

1. The problem statement, all variables and given/known data
Prove, for r $$\geq$$ 1,

Sr(n) = $$\sum$$ kr = (nr+1)/(r+1) + Pr(n)

where Pr(n) is some polynomial with respect to n of degree r.
(Hint: Prove using strong induction on r. After you show the base case is true, assume the statement istrue for all r = 1,2, ... ,R-1 and then show the statement is true for r = R.)

2. Relevant equations
(r + 1) $$\sum$$ kr = n(r+1) + Pr(n) - $$\sum$$ [c2kr-1 + c3kr-2 + ... + crk + 1], where c# is some constant.

3. The attempt at a solution
I tried to prove the base case (r=1), but I can't seem to complete it.
Using the equation given,
Sr(n) = $$\sum$$ kr = (nr+1)/(r+1)
-> the left side goes to: n2
And plugging in r=1, the right side goes to n2/2 + Pr(n)

Am I missing something? The two sides aren't equal at all, and I'm at a lost.