Rising Sum of Binomial Coefficients

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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