Prove sequence equivalence

Question

$$ \forall \text{ } n \in \mathbb{P} , \text{prove } 1*2 + 2 * 3 + ... + n (n+1) = \frac{1}{3}n(n+1)(n+2) $$

Proof via Induction

Base Case: n=1 $\implies 1(1+1)=\frac{1}{3}(1)(2)(3) \implies 2=2$

Induction: Assume that n=k and $1*2 + 2 * 3 + ... + k(k+1) = \frac{1}{3}k(k+1)(k+2)$ Show that this holds for k+1

$$ (k^{2}+k) + (k+1)(k+2) = \frac{1}{3} (k+1)(k+2)(k+3) $$ $$ 2(k+1)^{2} = \frac{1}{3}k^{3} +2(k+1)^{2} - \frac{1}{3}k$$

I'm so close here, but I know I'm missing something in making $k^{3}$ and $k$ equivalent on the right hand side of the equation. Any suggestions?

Here's a solution I posted to a related problem - perhaps you might find it helpful. http://math.stackexchange.com/questions/915793/telescoping-series-of-form-sum-n1-cdot-cdotnk/915881#915881 — Hypergeometricx, Oct 29 '14 at 17:20

score 1 · Answer 1 · answered Oct 29 '14 at 14:34

1

it must be $\frac{1}{3}n(n+1)(n+2)+(n+1)(n+2)=\frac{1}{3}(n+1)(n+2)(n+3)$

answered Oct 29 '14 at 14:34

1 Answers1