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.
$\blacksquare$