Sum of k Choose m up to n

Theorem

Let $m, n \in \Z: m \ge 0, n \ge 0$.

Then:

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

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

Proof

Proof by induction:

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

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

Basis for the Induction

$P(0)$ says $\displaystyle \binom 0 m = \binom 1 {m + 1}$.

When $m = 0$ we have by definition $\displaystyle \binom 0 0 = 1 = \binom 1 1$.

When $m > 0$ we also have by definition $\displaystyle \binom 0 m = 0 = \binom 1 {m+1}$.

This is our basis for the induction.

Induction Hypothesis

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

So this is our induction hypothesis:

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

Then we need to show:

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

Induction Step

This is our induction step:

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

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

Therefore:

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

$\blacksquare$

Alternative Proof

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \sum_{k=0}^n \binom k m\)	\(=\)	\(\displaystyle \sum_{0 \le m + k \le n} \binom {m + k} m\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sum_{-m \le k < 0} \binom {m + k} m + \sum_{0 \le k \le n - m} \binom {m + k} m\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 0 + \sum_{0 \le k \le n - m} \binom {m + k} m\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	as $\displaystyle \binom k m = 0$ for $k < 0, m \ge 0$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \binom {m + \left({n - m}\right) + 1} {n - m} = \binom {n + 1} {n - m}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	from Sum of r+k Choose k up to n‎
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \binom {n + 1} {\left ({n+1}\right) - \left({n - m}\right)}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Symmetry Rule for Binomial Coefficients‎
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \binom {n + 1} {m + 1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

$\blacksquare$