# Sum of Sequence of Fibonacci Numbers

From ProofWiki

## Contents

## Theorem

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

Then:

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

## Proof

From the initial definition of Fibonacci numbers, we have:

- $F_1 = 1, F_2 = 1, F_3 = 2, F_4 = 3$

Proof by induction:

For all $n \in \N_{>0}$, let $P \left({n}\right)$ be the proposition:

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

### Basis for the Induction

$P(2)$ is the case $F_1 + F_2 = 2 = F_4 - 1$, which holds.

This is our basis for the induction.

### Induction Hypothesis

Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:

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

Then we need to show:

- $\displaystyle \sum_{j \mathop = 1}^{k+1} F_j = F_{k+3} - 1$

### Induction Step

This is our induction step:

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

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\displaystyle F_{k+2} - 1 + F_{k+1}\) | \(\displaystyle \) | \(\displaystyle \) | by induction hypothesis | ||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\displaystyle F_{k+3} - 1\) | \(\displaystyle \) | \(\displaystyle \) | by definition of Fibonacci numbers |

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

Therefore:

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

$\blacksquare$

## Sources

- George E. Andrews:
*Number Theory*(1971)... (previous)... (next): $\S 1.1$: Exercise $7$