Rising Sum of Binomial Coefficients/Direct Proof

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

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

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

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

Direct Proof
This proof adds up all the terms of the summation to obtain the desired result.

Since the first term equals $1$, it may be replaced with $\displaystyle \binom {n+1} {n+1}$.

So:


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

The sum will be computed in $m$ steps, combining the first two terms with Pascal's Rule in each step.

Step 1:


 * $\displaystyle \binom {n + 1} {n + 1} + \binom {n + 1} n + \sum_{j \mathop = 2}^m \binom {n + j} n = \binom {n + 2} {n + 1} + \sum_{j \mathop = 2}^m \binom {n + j} n$

Step 2:


 * $\displaystyle \binom {n + 2} {n + 1} + \binom {n + 2} n + \sum_{j \mathop = 3}^m \binom {n + j} n = \binom {n + 3} {n + 1} + \sum_{j \mathop = 3}^m \binom {n + j} n$

After $m-1$ steps, we obtain:


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

Step $m$:


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

Hence the result.