# Sum of Sequence of n Choose 2/Proof 2

## Theorem

Let $n \in \Z$ be an integer such that $n > 2$.

 $\ds \sum_{j \mathop = 2}^n \dbinom j 2$ $=$ $\ds \dbinom 2 2 + \dbinom 3 2 + \dbinom 4 2 + \dotsb + \dbinom n 2$ $\ds$ $=$ $\ds \dbinom {n + 1} 3$

where $\dbinom n j$ denotes a binomial coefficient.

## Proof

The proof proceeds by induction.

For all $n \in \Z_{\ge 2}$, let $\map P n$ be the proposition:

$\displaystyle \sum_{j \mathop = 2}^n \dbinom j 2 = \dbinom {n + 1} 3$

### Basis for the Induction

$\map P 2$ is the case:

 $\ds \sum_{j \mathop = 2}^2 \dbinom j 2$ $=$ $\ds \dbinom 2 2$ $\ds$ $=$ $\ds 1$ $\ds$ $=$ $\ds \dbinom 2 2$

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:

$\sum_{j \mathop = 2}^k \dbinom j 2 = \dbinom {k + 1} 3$

from which it is to be shown that:

$\sum_{j \mathop = 2}^{k + 1} \dbinom j 2 = \dbinom {k + 2} 3$

### Induction Step

This is the induction step:

 $\ds \sum_{j \mathop = 2}^{k + 1} \dbinom j 2$ $=$ $\ds \sum_{j \mathop = 2}^k \dbinom j 2 + \dbinom {k + 1} 2$ $\ds$ $=$ $\ds \dbinom {k + 1} 3 + \dbinom {k + 1} 2$ Induction Hypothesis $\ds$ $=$ $\ds \dbinom {k + 2} 3$ Pascal's Rule

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

Therefore:

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