Rising Sum of Binomial Coefficients/Direct Proof

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:

$\displaystyle \binom n n + \binom {n+1} n + \binom {n+2} n + \cdots + \binom {n+m} n = \binom {n+m+1} {n+1} = \binom {n+m+1} m$

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:

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

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

Step 1:

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

Step 2:

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

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.

$\blacksquare$