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

Theorem
Let $\sequence {A_{n k} }$ 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_{\paren {n - 1} k} + A_{\paren {n - 1} \paren {k - 1} } + \dbinom n k & : \text{otherwise} \end{cases}$

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

Proof
The proof proceeds by induction on $n$.

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

Basis for the Induction
$\map P 0$ is the case $A_{0 k}$:

Let $k = 0$.

Then:

Let $k > 0$.

Then:

Thus $\map P 0$ is seen to hold.

This is the basis for the induction.

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

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

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

Induction Step
This is the induction step:

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

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