Sum of Even Sequence of Products of Consecutive Fibonacci Numbers

From ProofWiki
Jump to navigation Jump to search

Theorem

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

Then:

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


Proof

From Sum of Odd Sequence of Products of Consecutive Fibonacci Numbers:

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

Hence:

\(\ds \sum_{j \mathop = 1}^{2 n} F_j F_{j + 1}\) \(=\) \(\ds \sum_{j \mathop = 1}^{2 n - 1} F_j F_{j + 1} + F_{2 n} F_{2 n + 1}\)
\(\ds \) \(=\) \(\ds {F_{2 n} }^2 + F_{2 n} F_{2 n + 1}\) from $(1)$
\(\ds \) \(=\) \(\ds F_{2 n} \paren {F_{2 n} + F_{2 n + 1} }\)
\(\ds \) \(=\) \(\ds F_{2 n} F_{2 n + 2}\)
\(\ds \) \(=\) \(\ds {F_{2 n + 1} }^2 - 1\) Cassini's Identity

$\blacksquare$


Sources