Rising Sum of Binomial Coefficients/Direct Proof

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Theorem

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


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 $\dbinom {n + 1} {n + 1}$.

So:

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


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

Step 1:

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


Step 2:

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


After $m - 1$ steps, we obtain:

$\dbinom {n + m} {n + 1} + \dbinom {n + m} n$


Step $m$:

$\dbinom {n + m} {n + 1} + \dbinom {n + m} n = \dbinom {n + m + 1} {n + 1}$

Hence the result.

$\blacksquare$