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:

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


Proof

Proof by induction:

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

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


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:

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


Then we need to show:

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


Induction Step

This is our induction step:

\(\displaystyle \sum_{j \mathop = 1}^{2 k + 1} F_j F_{j + 1}\) \(=\) \(\displaystyle \sum_{j \mathop = 1}^{2 k - 1} F_j F_{j + 1} + F_{2 k} F_{2 k + 1} + F_{2 k + 1} F_{2 k + 2}\)
\(\displaystyle \) \(=\) \(\displaystyle {F_{2 k} }^2 + F_{2 k} F_{2 k + 1} + F_{2 k + 1} F_{2 k + 2}\) Induction hypothesis
\(\displaystyle \) \(=\) \(\displaystyle F_{2 k} \paren {F_{2 k} + F_{2 k + 1} } + F_{2 k + 1} F_{2 k + 2}\)
\(\displaystyle \) \(=\) \(\displaystyle F_{2 k} F_{2 k + 2} + F_{2 k + 1} F_{2 k + 2}\)
\(\displaystyle \) \(=\) \(\displaystyle F_{2 k + 2} \paren {F_{2 k} + F_{2 k + 1} }\)
\(\displaystyle \) \(=\) \(\displaystyle {F_{2 k + 2} }^2\)

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


Therefore:

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

$\blacksquare$


Sources