Rising Sum of Binomial Coefficients

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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