# Rising Sum of Binomial Coefficients

## Contents

## Theorem

Let $m, n \in \Z_{\ge 0}$ be positive integers.

Then:

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

where $\dbinom n k$ denotes a binomial coefficient.

That is:

- $\dbinom n n + \dbinom {n + 1} n + \dbinom {n + 2} n + \cdots + \dbinom {n + m} n = \dbinom {n + m + 1} {n + 1} = \dbinom {n + m + 1} m$

## Proof by Induction

Proof by induction:

Let $n \in \Z$.

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

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

$\map P 0$ is true, as this just says:

- $\dbinom n n = \dbinom {n + 1} {n + 1}$

But $\dbinom n n = \dbinom {n + 1} {n + 1} = 1$ from the Definition of Binomial Coefficient.

### Basis for the Induction

$\map P 1$ is the case:

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

\(\displaystyle \) | \(=\) | \(\displaystyle 1 + \paren {n + 1}\) | Definition of Binomial Coefficient | ||||||||||

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

\(\displaystyle \) | \(=\) | \(\displaystyle \binom {n + 2} {n + 1}\) | Definition of Binomial Coefficient |

So:

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

and $\map P 1$ is seen to hold.

This is our basis for the induction.

### Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ 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:

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

\(\displaystyle \) | \(=\) | \(\displaystyle \binom {n + k + 1} {n + 1} + \binom {n + k + 1} n\) | Induction Hypothesis | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \binom {n + k + 2} {n + 1}\) | Pascal's Rule |

So $\map P k \implies \map P {k + 1}$ 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:

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

$\blacksquare$

## 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 $\dbinom {n + 1} {n + 1}$.

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:

- $\dbinom {n + m} {n + 1} + \dbinom {n + m} n$

Step $m$:

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

Hence the result.

$\blacksquare$

## Proof 3

We have:

\(\displaystyle \binom {n + j} n\) | \(=\) | \(\displaystyle \binom {n + j} {\left({n + j}\right) - n}\) | Symmetry Rule for Binomial Coefficients | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \binom {n + j} j\) |

The result follows from Sum of $\dbinom {r + k} k$ up to $n$.

$\blacksquare$

## Marginal cases

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

### n = 0

When $n = 0$ we have:

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

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

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

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

So the theorem holds for $n = 0$.

$\blacksquare$

### n = 1

When $n = 1$ we have:

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

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

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

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

So the theorem holds for $n = 1$.

$\blacksquare$

## Sources

- 1968: Murray R. Spiegel:
*Mathematical Handbook of Formulas and Tables*... (previous) ... (next): $\S 3$: The Binomial Formula and Binomial Coefficients: $3.9$ - 1992: Larry C. Andrews:
*Special Functions of Mathematics for Engineers*... (previous) ... (next): $\S 1.2.4$: Factorials and binomial coefficients: $1.33$