Sum of k Choose m up to n

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

Then:
 * $$\sum_{k=1}^n \binom k m = \binom {n+1} {m+1}$$

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

Proof
Proof by induction:

For all $$n \in \N$$, let $$P \left({n}\right)$$ be the proposition:
 * $$\sum_{k=0}^n \binom k m = \binom {n+1} {m+1}$$.

Basis for the Induction
$$P(0)$$ says $$\binom 0 m = \binom 1 {m + 1}$$.

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

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

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:
 * $$\sum_{k=1}^r \binom k m = \binom {r+1} {m+1}$$.

Then we need to show:
 * $$\sum_{k=1}^{r+1} \binom k m = \binom {r+2} {m+1}$$.

Induction Step
This is our induction step:

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

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

Alternative Proof
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