Sum over k of n Choose k by Fibonacci t to the k by Fibonacci t-1 to the n-k by Fibonacci m+k

Theorem

 * $\ds \sum_{k \mathop \ge 0} \binom n k {F_t}^k {F_{t - 1} }^{n - k} F_{m + k} = F_{m + t n}$

where:
 * $\dbinom n k$ denotes a binomial coefficient
 * $F_n$ denotes the $n$th Fibonacci number.

Proof
The proof proceeds by induction on $n$.

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

$\map P 0$ is the case:


 * $\ds \binom 0 0 {F_t}^0 {F_{t - 1} }^0 F_{m + 0} = F_m$

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

Basis for the Induction
$\map P 1$ 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 $\map P q$ is true, where $q \ge 1$, then it logically follows that $\map P {q + 1}$ is true.

So this is the induction hypothesis:
 * $\forall m, t \in \N: \ds \sum_{k \mathop \ge 0} \binom q k {F_t}^k {F_{t - 1} }^{q - k} F_{m + k} = F_{m + t q}$

from which it is to be shown that:
 * $\forall m, t \in \N: \ds \sum_{k \mathop \ge 0} \binom {q + 1} k {F_t}^k {F_{t - 1} }^{q + 1 - k} F_{m + k} = F_{m + t \paren {q + 1} }$

Induction Step
This is the induction step:

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

Therefore:
 * $\forall m, n, t \in \N: \ds \sum_{k \mathop \ge 0} \binom n k {F_t}^k {F_{t - 1} }^{n - k} F_{m + k} = F_{m + t n}$

Also see
The following are corollaries to this theorem:


 * Sum over k of n Choose k by Fibonacci Number with index m+k


 * Sum of Sequence of Product of Fibonacci Number with Binomial Coefficient