# Sum of Sequence of n Choose 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.

### Corollary

Let $n \in \Z_{>0}$ be a (strictly) positive integer.

Let $T_n$ denote the $n$th triangular number.

Then:

\(\ds \sum_{j \mathop = 1}^n T_j\) | \(=\) | \(\ds T_1 + T_2 + T_3 + \dotsb + T_n\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \dfrac {n \paren {n + 1} \paren {n + 2} } 6\) |

## Proof 1

We can rewrite the left hand side as:

- $\displaystyle \sum_{j \mathop = 0}^m \dbinom {2 + j} 2$

where $m = n - 2$.

From Rising Sum of Binomial Coefficients:

- $\displaystyle \sum_{j \mathop = 0}^m \binom {n + j} n = \binom {n + m + 1} {n + 1}$

The result follows by setting $n = 2$ and changing the upper index from $m$ to $n - 2$.

$\blacksquare$

## Proof 2

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:

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