Rising Sum of Binomial Coefficients/Proof by Induction

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Theorem

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


Proof

Proof by induction:

Let $n \in \Z$.

For all $m \in \N$, let $\map P m$ be the proposition:

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


$\map P 0$ 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:

\(\ds \sum_{j \mathop = 0}^1 \binom {n + j} n\) \(=\) \(\ds \binom n n + \binom {n + 1} n\)
\(\ds \) \(=\) \(\ds 1 + \paren {n + 1}\) Definition of Binomial Coefficient
\(\ds \) \(=\) \(\ds n + 2\)
\(\ds \) \(=\) \(\ds \binom {n + 2} {n + 1}\) Definition of Binomial Coefficient

So:

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

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:

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


Then we need to show:

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


Induction Step

This is our induction step:

\(\ds \sum_{j \mathop = 0}^{k + 1} \binom {n + j} n\) \(=\) \(\ds \sum_{j \mathop = 0}^k \binom {n + j} n + \binom {n + k + 1} n\)
\(\ds \) \(=\) \(\ds \binom {n + k + 1} {n + 1} + \binom {n + k + 1} n\) Induction Hypothesis
\(\ds \) \(=\) \(\ds \binom {n + k + 2} {n + 1}\) Pascal's Rule

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.


Therefore:

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


Finally, from Symmetry Rule for Binomial Coefficients:

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

$\blacksquare$


Proof

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