Sum of Sequence of Products of Consecutive Integers/Proof 1

Theorem

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

Proof
Proof by induction:

For all $n \in \N_{>0}$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle \forall n \ge 1: \sum_{j \mathop = 1}^n j \left({j+1}\right) = \frac {n \left({n+1}\right) \left({n+2}\right)} 3$

Basis for the Induction
$P(1)$ is true, as this just says $1 \times 2 = 2 = \dfrac {1 \times 2 \times 3} 3$.

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:
 * $\displaystyle \sum_{j \mathop = 1}^k j \left({j+1}\right) = \frac {k \left({k+1}\right) \left({k+2}\right)} 3$

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

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