Sum of Sequence of Products of Consecutive Fibonacci Numbers

Theorem

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

Then:

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

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

Proof

Proof by induction:

For all $n \in \N^*$, let $P \left({n}\right)$ be the proposition:

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

Basis for the Induction

• $P(1)$ 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 $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:

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

Then we need to show:

$\displaystyle \sum_{j = 1}^{2k+1} F_j F_{j+1} = F_{2 \left({k + 1}\right)}^{\ 2}$

Induction Step

This is our induction step:

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

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

Therefore:

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

For the second result, we have:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \sum_{j = 1}^{2n} F_j F_{j+1}\)	\(=\)	\(\displaystyle \sum_{j = 1}^{2n-1} F_j F_{j+1} + F_{2n} F_{2n+1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle F_{2n}^{\ 2} + F_{2n} F_{2n+1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	from above
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle F_{2n} \left({F_{2n} + F_{2n+1} }\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle F_{2n} F_{2n + 2}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle F_{2n+1}^{\ 2} - 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Cassini's Identity

$\blacksquare$

Sources

• George E. Andrews: Number Theory (1971): $\S 1.1$: Exercise $11, \ 12$