Sum of r+k Choose k up to n

Theorem

Let $r \in \R$ be a real number.

Then:

$\displaystyle \forall n \in \Z: n \ge 0: \sum_{k=0}^n \binom {r+k} k = \binom {r + n + 1} n$

where $\displaystyle \binom r k$ is a binomial coefficient.

Proof

Proof by induction:

For all $n \in \Z: n \ge 0$, let $P \left({n}\right)$ be the proposition

$\displaystyle \sum_{k=0}^n \binom {r+k} k = \binom {r + n + 1} n$

Basis for the Induction

$P(0)$ is true, as $\displaystyle \binom r 0 = 1 = \binom {r+1} 0$ by definition.

This is our basis for the induction.

Induction Hypothesis

• Now we need to show that, if $P \left({m}\right)$ is true, where $r \ge 2$, then it logically follows that $P \left({m+1}\right)$ is true.

So this is our induction hypothesis:

$\displaystyle \sum_{k=0}^m \binom {r+k} k = \binom {r + m + 1} m$

Then we need to show:

$\displaystyle \sum_{k=0}^{m+1} \binom {r+k} k = \binom {r + m + 2} {m + 1}$

Induction Step

This is our induction step:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \sum_{k=0}^{m+1} \binom {r+k} k\)	\(=\)	\(\displaystyle \sum_{k=0}^m \binom {r+k} k + \binom {r + m + 1} {m + 1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \binom {r + m + 1} m + \binom {r + m + 1} {m + 1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	from the induction hypothesis
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \binom {r + m + 2} {m + 1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	from Pascal's Rule

So $P \left({m}\right) \implies P \left({m+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:

$\displaystyle \forall n \in \Z: n \ge 0: \sum_{k=0}^n \binom {r+k} k = \binom {r + n + 1} n$

$\blacksquare$