Binomial Coefficient n Choose k by n Plus 1 Minus n Choose k + 1

Theorem
Let $\left\langle{A_{n k} }\right\rangle$ be a sequence defined on $n, k \in \Z_{\ge 0}$ as:


 * $A_{n k} = \begin{cases} 1 & : k = 0 \\

0 & : k \ne 0, n = 0 \\ A_{\left({n - 1}\right) k} + A_{\left({n - 1}\right) \left({k - 1}\right)} + \dbinom n k & : \text{otherwise} \end{cases}$

Then the closed form for $A_{n k}$ is given as:
 * $A_{n k} = \left({n + 1}\right) \dbinom n k - \dbinom n {k + 1}$

Proof
The proof proceeds by induction on $n$.

For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition:
 * $A_{n k} = \left({n + 1}\right) \dbinom n k - \dbinom n {k + 1} + \dbinom n k$

Basis for the Induction
$P \left({0}\right)$ is the case $A_{0 k}$:

Let $k = 0$.

Then:

Let $k > 0$.

Then:

Thus $P \left({0}\right)$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $P \left({r}\right)$ is true, where $r \ge 0$, then it logically follows that $P \left({r + 1}\right)$ is true.

So this is the induction hypothesis:
 * $A_{r k} = \left({r + 1}\right) \dbinom r k - \dbinom r {k + 1}$

from which it is to be shown that:
 * $A_{\left({r + 1}\right) k} = \left({r + 2}\right) \dbinom {r + 1} k - \dbinom {r + 1} {k + 1}$

Induction Step
This is the induction step:

So $P \left({r}\right) \implies P \left({r + 1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\forall n, k \in \Z_{\ge 0}: A_{n k} = \left({n + 1}\right) \dbinom n k - \dbinom n {k + 1}$