Rising Sum of Binomial Coefficients

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:

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

Corollary

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

That is:

$\displaystyle \binom n 0 + \binom {n+1} 1 + \binom {n+2} 2 + \cdots + \binom {n+m} m = \binom {n+m+1} m$

Proof by Induction

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.

$\blacksquare$

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 $\displaystyle \binom {n+1} {n+1}$

So:

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

The sum will be computed in $m$ steps, combining the first two terms with Pascal's Rule in each step.

Step 1:

$\displaystyle \binom {n+1} {n+1} + \binom {n+1} n + \sum_{j=2}^m \binom {n + j} n = \binom {n+2} {n+1} + \sum_{j=2}^m \binom {n + j} n$

Step 2:

$\displaystyle \binom {n+2} {n+1} + \binom {n+2} n + \sum_{j=3}^m \binom {n + j} n = \binom {n+3} {n+1} + \sum_{j=3}^m \binom {n + j} n$

After $m-1$ steps, we obtain:

$\displaystyle \binom {n+m} {n+1} + \binom {n+m} {n}$

Step $m$:

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

Hence the result.

$\blacksquare$

Marginal cases

Just to make sure, it is worth checking the marginal cases:

n = 0

When $n = 0$ we have:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \sum_{j=0}^m \binom j 0\)	\(=\)	\(\displaystyle \binom 0 0 + \binom 1 0 + \binom 2 0 + \cdots \binom m 0\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 1 + 1 + \cdots + 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	from $0$ to $m$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle m + 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	as there are $m+1$ of them
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \binom {m+1} 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

So the theorem holds for $n = 0$.

$\blacksquare$

n = 1

When $n = 1$ we have:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \sum_{j=0}^m \binom {1 + j} 1\)	\(=\)	\(\displaystyle \binom 1 1 + \binom 2 1 + \binom 3 1 + \cdots \binom {m + 1} 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 1 + 2 + \cdots + \left({m + 1}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {\left({m + 1}\right) \left({m + 2}\right)} 2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Closed Form for Triangular Numbers
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \binom {m+2} 2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Closed Form for Triangular Numbers

So the theorem holds for $n = 1$.

$\blacksquare$

Proof of Corollary

From the Symmetry Rule for Binomial Coefficients we have:

$\displaystyle \binom {n + j} n = \binom {n + j} j$

and

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

Hence the result.

$\blacksquare$

Sources

• Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (1968): $\S 1.2.6 \ \text{E}$
• Murray R. Spiegel: Mathematical Handbook of Formulas and Tables (1968): $3.9$
• George E. Andrews: Number Theory (1971): $\S 3.1$: Exercise $5$