Rising Sum of Binomial Coefficients/Proof 3

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 $n$.

$\blacksquare$