Rising Sum of Binomial Coefficients

Theorem
Let $$n \in \Z$$ be an integer such that $$n \ge 0$$.

Then:


 * $$\sum_{j=0}^m \binom {n + j} n = \binom {n+m+1} {n+1}$$

where $$\binom n k$$ denotes a binomial coefficient.

That is:
 * $$\binom n n + \binom {n+1} n + \binom {n+2} n + \cdots + \binom {n+m} n = \binom {n+m+1} {n+1}$$

Proof
Proof by induction:

Let $$n \in \Z$$.

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

$$P(0)$$ is true, as this just says $$\binom n n = \binom {n+1} {n+1}$$.

But $$\binom n n = \binom {n+1} {n+1} = 1$$ from the definition of a binomial coefficient.

Basis for the Induction
$$P(1)$$ is the case:

$$ $$ $$ $$

So:
 * $$\sum_{j=0}^1 \binom {n + j} n = \binom {n + 2} {n + 1}$$ and $$P(1)$$ is seen to hold.

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:
 * $$\sum_{j=0}^k \binom {n + j} n = \binom {n+k+1} {n+1}$$

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

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:
 * $$\sum_{j=0}^m \binom {n + j} n = \binom {n+m+1} {n+1}$$

Marginal cases
Just to make sure, it is worth checking the marginal cases:

n = 0
When $$n = 0$$ we have:

$$ $$ $$ $$

So the theorem holds for $$n = 0$$.

n = 1
When $$n = 1$$ we have:

$$ $$ $$ $$

So the theorem holds for $$m = 1$$.