Sum of Sequence of n Choose 2/Proof 2

Proof
The proof proceeds by induction.

For all $n \in \Z_{\ge 2}$, let $\map P n$ be the proposition:
 * $\ds \sum_{j \mathop = 2}^n \dbinom j 2 = \dbinom {n + 1} 3$

Basis for the Induction
$\map P 2$ is the case:

Thus $\map P 2$ 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 $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.

So this is the induction hypothesis:
 * $\ds \sum_{j \mathop = 2}^k \dbinom j 2 = \dbinom {k + 1} 3$

from which it is to be shown that:
 * $\ds \sum_{j \mathop = 2}^{k + 1} \dbinom j 2 = \dbinom {k + 2} 3$

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:
 * $\forall n \in \Z_{\ge 2}: \ds \sum_{j \mathop = 2}^n \dbinom j 2 = \dbinom {n + 1} 3$