Fibonacci Number plus Binomial Coefficient in terms of Fibonacci Numbers

Theorem
Let $m \in \Z_{>0}$ be a positive integer.

Let $\sequence {a_n}$ be the sequence defined as:
 * $a_n = \begin{cases}

0 & : n = 0 \\ 1 & : n = 1 \\ a_{n - 2} + a_{n - 1} + \dbinom {n - 2} m & : n > 1 \end{cases}$

where $\dbinom {n - 2} m$ denotes a binonial coefficient.

Then $\sequence {a_n}$ can be expressed in Fibonacci numbers as:
 * $a_n = F_{m + 1} F_{n - 1} + \paren {F_{m + 2} + 1} F_n - \ds \sum_{k \mathop = 0}^m \binom {n + m - k} k$

Proof
The proof proceeds by induction.

For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
 * $a_n = F_{m + 1} F_{n - 1} + \paren {F_{m + 2} + 1} F_n - \ds \sum_{k \mathop = 0}^m \binom {n + m - k} k$

Basis for the Induction
$\map P 0$ is the case:

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

$\map P 1$ is the case:

Thus $\map P 1$ is seen to hold.

This is the basis for the induction.

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

So this is the induction hypothesis:
 * $a_{k - 1} = F_{m + 1} F_{k - 2} + \paren {F_{m + 2} + 1} F_{k - 1} - \ds \sum_{r \mathop = 0}^m \binom {k - 1 + m - r} r$
 * $a_k = F_{m + 1} F_{k - 1} + \paren {F_{m + 2} + 1} F_k - \ds \sum_{r \mathop = 0}^m \binom {k + m - r} r$

from which it is to be shown that:
 * $a_{k + 1} = F_{m + 1} F_k + \paren {F_{m + 2} + 1} F_{k + 1} - \ds \sum_{r \mathop = 0}^m \binom {k + 1 + m - r} r$

Induction Step
This is the induction step:

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

Therefore:
 * $\forall n \in \Z_{\ge 0}: a_n = F_{n - 1} r + F_n s$