Sum of Sequence of Products of 3 Consecutive Integers

Proof
Proof by induction:

For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
 * $\ds \forall n \ge 1: \sum_{j \mathop = 1}^n j \paren {j + 1} \paren {j + 2} = \dfrac {n \paren {n + 1} \paren {n + 2} \paren {n + 3} } 4$

Basis for the Induction
$\map P 1$ is the case:

Thus $\map P 1$ is seen to be true.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:
 * $\ds \sum_{j \mathop = 1}^k j \paren {j + 1} \paren {j + 2} = \dfrac {k \paren {k + 1} \paren {k + 2} \paren {k + 3} } 4$

Then we need to show:
 * $\ds \sum_{j \mathop = 1}^{k + 1} j \paren {j + 1} \paren {j + 2} = \dfrac {\paren {k + 1} \paren {k + 2} \paren {k + 3} \paren {k + 4} } 4$

Induction Step
This is our induction step:

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

Therefore:
 * $\ds \forall n \ge 1: \sum_{j \mathop = 1}^n j \paren {j + 1} \paren {j + 2} = \dfrac {n \paren {n + 1} \paren {n + 2} \paren {n + 3} } 4$