Binomial Coefficient with Two
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Theorem
- $\forall r \in \R: \dbinom r 2 = \dfrac {r \paren {r - 1} } 2$
Corollary
The usual presentation of this result is:
- $\forall n \in \N: \dbinom n 2 = T_{n - 1} = \dfrac {n \paren {n - 1} } 2$
where $T_n$ is the $n$th triangular number.
Proof
From the definition of binomial coefficients:
- $\dbinom r k = \dfrac {r^{\underline k}} {k!}$ for $k \ge 0$
where $r^{\underline k}$ is the falling factorial.
In turn:
- $\ds x^{\underline k} := \prod_{j \mathop = 0}^{k - 1} \paren {x - j}$
When $k = 2$:
- $\ds \prod_{j \mathop = 0}^1 \paren {x - j} = \frac {\paren {x - 0} \paren {x - 1} } {2!}$
where $2! = 1 \times 2 = 2$.
So:
- $\forall r \in \R: \dbinom r 2 = \dfrac {r \paren {r - 1} } 2$
$\blacksquare$
Also see
- Particular Values of Binomial Coefficients for other similar results.
Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.6$: Binomial Coefficients: $(4)$