# Rising Sum of Binomial Coefficients

## Contents

## Theorem

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

Then:

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

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 by Induction

Proof by induction:

Let $n \in \Z$.

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

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

$P(0)$ is true, as this just says:

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

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:

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \sum_{j \mathop = 0}^1 \binom {n + j} n\) | \(=\) | \(\displaystyle \) | \(\displaystyle \binom n n + \binom {n + 1} n\) | \(\displaystyle \) | \(\displaystyle \) | |||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\displaystyle 1 + \left({n + 1}\right)\) | \(\displaystyle \) | \(\displaystyle \) | from the definition of a binomial coefficient | ||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\displaystyle n + 2\) | \(\displaystyle \) | \(\displaystyle \) | |||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\displaystyle \binom {n + 2} {n + 1}\) | \(\displaystyle \) | \(\displaystyle \) | from the definition of a binomial coefficient |

So:

- $\displaystyle \sum_{j \mathop = 0}^1 \binom {n + j} n = \binom {n + 2} {n + 1}$ 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:

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

Then we need to show:

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

### Induction Step

This is our induction step:

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \sum_{j \mathop = 0}^{k + 1} \binom {n + j} n\) | \(=\) | \(\displaystyle \) | \(\displaystyle \sum_{j \mathop = 0}^k \binom {n + j} n + \binom {n + k + 1} n\) | \(\displaystyle \) | \(\displaystyle \) | |||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\displaystyle \binom {n + k + 1} {n + 1} + \binom {n + k + 1} n\) | \(\displaystyle \) | \(\displaystyle \) | from the induction hypothesis | ||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\displaystyle \binom {n + k + 2} {n + 1}\) | \(\displaystyle \) | \(\displaystyle \) | 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:

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

Finally, from Symmetry Rule for Binomial Coefficients:

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

$\blacksquare$

## 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:

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

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

Step 1:

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

Step 2:

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

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:

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \sum_{j \mathop = 0}^m \binom j 0\) | \(=\) | \(\displaystyle \) | \(\displaystyle \binom 0 0 + \binom 1 0 + \binom 2 0 + \cdots \binom m 0\) | \(\displaystyle \) | \(\displaystyle \) | |||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\displaystyle 1 + 1 + \cdots + 1\) | \(\displaystyle \) | \(\displaystyle \) | from $0$ to $m$ | ||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\displaystyle m + 1\) | \(\displaystyle \) | \(\displaystyle \) | as there are $m+1$ of them | ||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\displaystyle \binom {m+1} 1\) | \(\displaystyle \) | \(\displaystyle \) |

So the theorem holds for $n = 0$.

$\blacksquare$

### n = 1

When $n = 1$ we have:

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \sum_{j \mathop = 0}^m \binom {1 + j} 1\) | \(=\) | \(\displaystyle \) | \(\displaystyle \binom 1 1 + \binom 2 1 + \binom 3 1 + \cdots \binom {m + 1} 1\) | \(\displaystyle \) | \(\displaystyle \) | |||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\displaystyle 1 + 2 + \cdots + \left({m + 1}\right)\) | \(\displaystyle \) | \(\displaystyle \) | |||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\displaystyle \frac {\left({m + 1}\right) \left({m + 2}\right)} 2\) | \(\displaystyle \) | \(\displaystyle \) | Closed Form for Triangular Numbers | ||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\displaystyle \binom {m+2} 2\) | \(\displaystyle \) | \(\displaystyle \) | Closed Form for Triangular Numbers |

So the theorem holds for $n = 1$.

$\blacksquare$

## 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)... (previous)... (next): $\S 3$: The Binomial Formula and Binomial Coefficients: $3.9$ - George E. Andrews:
*Number Theory*(1971)... (previous)... (next): $\S 3.1$: Exercise $5$ - Larry C. Andrews:
*Special Functions of Mathematics for Engineers*(1992)... (previous)... (next): $\S 1.2.4$: Factorials and binomial coefficients: $1.33$