Rising Sum of Binomial Coefficients/Proof by Induction

Theorem

Let $n \in \Z$ be an integer such that $n \ge 0$.

Then:

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

where $\displaystyle \binom n k$ denotes a binomial coefficient.

That is:

$\displaystyle \binom n n + \binom {n+1} n + \binom {n+2} n + \cdots + \binom {n+m} n = \binom {n+m+1} {n+1} = \binom {n+m+1} m$

Proof

Proof by induction:

Let $n \in \Z$.

For all $m \in \N$, let $P \left({m}\right)$ be the proposition:

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

$P(0)$ is true, as this just says $\binom n n = \binom {n+1} {n+1}$.

But $\displaystyle \binom n n = \binom {n+1} {n+1} = 1$ from the definition of a binomial coefficient.

Basis for the Induction

$P(1)$ is the case:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \sum_{j=0}^1 \binom {n + j} n\)	\(=\)	\(\displaystyle \binom n n + \binom {n + 1} n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 1 + \left({n+1}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	from the definition of a binomial coefficient
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle n + 2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \binom {n + 2} {n + 1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	from the definition of a binomial coefficient

So:

$\displaystyle \sum_{j=0}^1 \binom {n + j} n = \binom {n + 2} {n + 1}$ and $P(1)$ is seen to hold.

This is our basis for the induction.

Induction Hypothesis

Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:

$\displaystyle \sum_{j=0}^k \binom {n + j} n = \binom {n+k+1} {n+1}$

Then we need to show:

$\displaystyle \sum_{j=0}^{k+1} \binom {n + j} n = \binom {n+k+2} {n+1}$.

Induction Step

This is our induction step:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \sum_{j=0}^{k+1} \binom {n + j} n\)	\(=\)	\(\displaystyle \sum_{j=0}^k \binom {n + j} n + \binom {n + k + 1} n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \binom {n+k+1} {n+1} + \binom {n + k + 1} n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	from the induction hypothesis
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \binom {n+k+2} {n+1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by Pascal's Rule

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:

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

Finally, we note that $\displaystyle \binom {n+m+1} {n+1} = \binom {n+m+1} m$ from Symmetry Rule for Binomial Coefficients.

$\blacksquare$