Sum of Sequence of Fibonacci Numbers

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 $7$