Rising Sum of Binomial Coefficients/Marginal Cases

Theorem

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

$n = 0$

When $n = 0$ we have:

\(\ds \sum_{j \mathop = 0}^m \binom j 0\)

\(=\)

\(\ds \binom 0 0 + \binom 1 0 + \binom 2 0 + \cdots \binom m 0\)

\(\ds \)

\(=\)

\(\ds 1 + 1 + \cdots + 1\)

from $0$ to $m$

\(\ds \)

\(=\)

\(\ds m + 1\)

as there are $m + 1$ of them

\(\ds \)

\(=\)

\(\ds \binom {m + 1} 1\)

So the theorem holds for $n = 0$.

$\Box$

$n = 1$

When $n = 1$ we have:

\(\ds \sum_{j \mathop = 0}^m \binom {1 + j} 1\)

\(=\)

\(\ds \binom 1 1 + \binom 2 1 + \binom 3 1 + \cdots \binom {m + 1} 1\)

\(\ds \)

\(=\)

\(\ds 1 + 2 + \cdots + \paren {m + 1}\)

\(\ds \)

\(=\)

\(\ds \frac {\paren {m + 1} \paren {m + 2} } 2\)

Closed Form for Triangular Numbers

\(\ds \)

\(=\)

\(\ds \binom {m + 2} 2\)

Definition of Binomial Coefficient

So the theorem holds for $n = 1$.

$\Box$

From Symmetry Rule for Binomial Coefficients:

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

thus:

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

and:

$\dbinom {m + 2} 2 = \dbinom {m + 2} m$

$\blacksquare$