Sum of r+k Choose k up to n

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

Then:
 * $$\forall n \in \Z: n \ge 0: \sum_{j=0}^n \binom {r+j} j = \binom {r + n + 1} n$$

where $$\binom r j$$ is a binomial coefficient.

Proof
Proof by induction:

For all $$n \in \Z: n \ge 0$$, let $$P \left({n}\right)$$ be the proposition
 * $$\sum_{j=0}^n \binom {r+j} j = \binom {r + n + 1} n$$

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

This is our basis for the induction.

Induction Hypothesis

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

So this is our induction hypothesis:
 * $$\sum_{j=0}^k \binom {r+j} j = \binom {r + k + 1} k$$

Then we need to show:
 * $$\sum_{j=0}^{k+1} \binom {r+j} j = \binom {r + k + 2} {k + 1}$$

Induction Step
This is our induction step:

$$ $$ $$

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

Therefore:
 * $$\forall n \in \Z: n \ge 0: \sum_{j=0}^n \binom {r+j} j = \binom {r + n + 1} n$$