Rising Sum of Binomial Coefficients

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

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

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

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

Corollary

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

That is:
 * $\displaystyle \binom n 0 + \binom {n+1} 1 + \binom {n+2} 2 + \cdots + \binom {n+m} m = \binom {n+m+1} m$

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 $n = 1$.

Proof of Corollary
From the Symmetry Rule for Binomial Coefficients we have:
 * $\displaystyle \binom {n + j} n = \binom {n + j} j$

and
 * $\displaystyle \binom {n+m+1} {n+1} = \binom {n+m+1} m$

Hence the result.