Sum of Sequence of n Choose 2/Proof 1

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Theorem

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

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

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


Proof

We can rewrite the left hand side as:

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

where $m = n - 2$.

From Rising Sum of Binomial Coefficients:

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


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

$\blacksquare$