Sum of Sequence of Odd Index Fibonacci Numbers

From ProofWiki
Jump to navigation Jump to search

Theorem

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


Then:

\(\, \displaystyle \forall n \ge 1: \, \) \(\displaystyle \sum_{j \mathop = 1}^n F_{2 j - 1}\) \(=\) \(\displaystyle F_1 + F_3 + F_5 + \cdots + F_{2 n - 1}\)
\(\displaystyle \) \(=\) \(\displaystyle F_{2 n}\)


Proof

Proof by induction:

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

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


Basis for the Induction

$\map P 1$ is the case $F_1 = 1 = F_2$, which holds from the definition of Fibonacci numbers.

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:

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


Then we need to show:

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


Induction Step

This is our induction step:

\(\displaystyle \sum_{j \mathop = 1}^{k + 1} F_{2 j - 1}\) \(=\) \(\displaystyle \sum_{j \mathop = 1}^k F_{2 j - 1} + F_{2 k + 1}\)
\(\displaystyle \) \(=\) \(\displaystyle F_{2 k} + F_{2 k + 1}\) Induction Hypothesis
\(\displaystyle \) \(=\) \(\displaystyle F_{2 k + 2}\) Definition of Fibonacci Numbers

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


Therefore:

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

$\blacksquare$


Sources