# Sum of Odd Sequence of Products of Consecutive Fibonacci Numbers

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## Theorem

Let $F_k$ be the $k$'th Fibonacci number.

Then:

- $\ds \sum_{j \mathop = 1}^{2 n - 1} F_j F_{j + 1} = {F_{2 n} }^2$

## Proof

Proof by induction:

For all $n \in \N_{>0}$, let $\map P n$ be the proposition:

- $\ds \sum_{j \mathop = 1}^{2 n - 1} F_j F_{j + 1} = {F_{2 n} }^2$

### Basis for the Induction

$\map P 1$ is true, as this just says $F_1 F_2 = 1 \times 1 = 1 = {F_2}^2$.

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:

- $\ds \sum_{j \mathop = 1}^{2 k - 1} F_j F_{j + 1} = {F_{2 k} }^2$

Then we need to show:

- $\ds \sum_{j \mathop = 1}^{2 k + 1} F_j F_{j + 1} = {F_{2 \paren {k + 1} } }^2$

### Induction Step

This is our induction step:

\(\ds \sum_{j \mathop = 1}^{2 k + 1} F_j F_{j + 1}\) | \(=\) | \(\ds \sum_{j \mathop = 1}^{2 k - 1} F_j F_{j + 1} + F_{2 k} F_{2 k + 1} + F_{2 k + 1} F_{2 k + 2}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds {F_{2 k} }^2 + F_{2 k} F_{2 k + 1} + F_{2 k + 1} F_{2 k + 2}\) | Induction hypothesis | |||||||||||

\(\ds \) | \(=\) | \(\ds F_{2 k} \paren {F_{2 k} + F_{2 k + 1} } + F_{2 k + 1} F_{2 k + 2}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds F_{2 k} F_{2 k + 2} + F_{2 k + 1} F_{2 k + 2}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds F_{2 k + 2} \paren {F_{2 k} + F_{2 k + 1} }\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds {F_{2 k + 2} }^2\) |

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

Therefore:

- $\ds \forall n \ge 1: \sum_{j \mathop = 1}^{2 n - 1} F_j F_{j + 1} = {F_{2 n} }^2$

$\blacksquare$

## Sources

- 1971: George E. Andrews:
*Number Theory*... (previous) ... (next): $\text {1-1}$ Principle of Mathematical Induction: Exercise $11$