Rising Sum of Binomial Coefficients/Proof 3
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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
We have:
\(\ds \binom {n + j} n\) | \(=\) | \(\ds \binom {n + j} {\left({n + j}\right) - n}\) | Symmetry Rule for Binomial Coefficients | |||||||||||
\(\ds \) | \(=\) | \(\ds \binom {n + j} j\) |
The result follows from Sum of $\dbinom {r + k} k$ up to $n$.
$\blacksquare$