Binomial Coefficient with Two

Theorem

 * $\displaystyle \forall r \in \R: \binom r 2 = \frac {r \left({r - 1}\right)} 2$

The usual presentation of this result is:
 * $\displaystyle \forall n \in \N: \binom n 2 = T_n = \frac {n \left({n - 1}\right)} 2$

where $T_n$ is the $n$th triangular number.

Proof
From the definition of binomial coefficients:


 * $\displaystyle \binom r k = \dfrac {r^{\underline k}} {k!}$ for $k \ge 0$

where $r^{\underline k}$ is the falling factorial.

In turn:


 * $\displaystyle x^{\underline k} := \prod_{j=0}^{k-1} \left({x - j}\right)$

When $k = 2$, we have:
 * $\displaystyle \prod_{j=0}^1 \left({x - j}\right) = \frac {\left({x - 0}\right) \left({x - 1}\right)} {2!}$

where $2! = 1 \times 2 = 2$.

So:


 * $\displaystyle \forall r \in \R: \binom r 2 = \frac {r \left({r - 1}\right)} 2$

This is completely compatible with the result for natural numbers:
 * $\displaystyle \forall n \in \N: \binom n 2 = \frac {n \left({n - 1}\right)} 2$

as from the definition:
 * $\displaystyle \binom n 2 = \dfrac {n!} {2! \ \left({n - 2}\right)!}$

the result following directly, again from the definition of the factorial where $2! = 1 \times 2$.

This is also proved in Closed Form for Triangular Numbers: Proof using Binomial Coefficients.

Also see

 * Particular Values of Binomial Coefficients for other similar results.