Sum of Sequence of Products of Consecutive Integers

Theorem

 * $$\sum_{j = 1}^n j \left({j+1}\right) = \frac {n \left({n+1}\right) \left({n+2}\right)} 3$$

Proof
Proof by induction:

For all $$n \in \N^*$$, let $$P \left({n}\right)$$ be the proposition $$\forall n \ge 1: \sum_{j = 1}^n j \left({j+1}\right) = \frac {n \left({n+1}\right) \left({n+2}\right)} 3$$.

Basis for the Induction

 * $$P(1)$$ is true, as this just says $$1 \times 2 = 2 = \frac {1 \times 2 \times 3} 3$$.

This is our basis for the induction.

Induction Hypothesis

 * Now we need to show that, if $$P \left({k}\right)$$ is true, where $$k \ge 2$$, then it logically follows that $$P \left({k+1}\right)$$ is true.

So this is our induction hypothesis:
 * $$\sum_{j = 1}^k j \left({j+1}\right) = \frac {k \left({k+1}\right) \left({k+2}\right)} 3$$.

Then we need to show:
 * $$\sum_{j = 1}^{k+1} j \left({j+1}\right) = \frac {\left({k+1}\right) \left({k+2}\right) \left({k+3}\right) } 3$$.

Induction Step
This is our induction step:

$$ $$ $$ $$

So $$P \left({k}\right) \implies P \left({k+1}\right)$$ and the result follows by the Principle of Mathematical Induction.

Therefore $$\forall n \ge 1: \sum_{j = 1}^n j \left({j+1}\right) = \frac {n \left({n+1}\right) \left({n+2}\right)} 3$$.