Sum of Sequence of Products of Consecutive Odd and Consecutive Even Numbers

Proof
The proof proceeds by induction.

For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
 * $\displaystyle \sum_{j \mathop = 1}^n j \paren {j + 2} = \frac {n \paren {n + 1} \paren {2 n + 7} } 6$

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

Thus $\map P 1$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown 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 the induction hypothesis:
 * $\displaystyle \sum_{j \mathop = 1}^k j \paren {j + 2} = \frac {k \paren {k + 1} \paren {2 k + 7} } 6$

from which it is to be shown that:
 * $\displaystyle \sum_{j \mathop = 1}^{k + 1} j \paren {j + 2} = \frac {\paren {k + 1} \paren {k + 2} \paren {2 \paren {k + 1} + 7} } 6$

Induction Step
This is the induction step:

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

Therefore:
 * $\displaystyle \forall n \in \Z_{>0}: \sum_{j \mathop = 1}^n j \paren {j + 2} = \frac {n \paren {n + 1} \paren {2 n + 7} } 6$