Sum of Sequence of Fibonacci Numbers

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Theorem

Let $F_n$ denote the $n$th Fibonacci number.


Then:

$\displaystyle \forall n \in \Z_{\ge 0}: \sum_{j \mathop = 0}^n F_j = F_{n + 2} - 1$


Proof

Proof by induction:

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

$\displaystyle \sum_{j \mathop = 0}^n F_j = F_{n + 2} - 1$


$\map P 0$ is the case:

\(\displaystyle F_0\) \(=\) \(\displaystyle 0\)
\(\displaystyle \) \(=\) \(\displaystyle 1 - 1\)
\(\displaystyle \) \(=\) \(\displaystyle F_2 - 1\)

which is seen to hold.


Basis for the Induction

$\map P 1$ is the case:

\(\displaystyle F_1\) \(=\) \(\displaystyle 1\)
\(\displaystyle \) \(=\) \(\displaystyle 2 - 1\)
\(\displaystyle \) \(=\) \(\displaystyle F_3 - 1\)

which is seen to hold.

This is our basis for the induction.


Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ 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:

\(\displaystyle \sum_{j \mathop = 1}^{k + 1} F_j\) \(=\) \(\displaystyle \sum_{j \mathop = 1}^k F_j + F_{k + 1}\)
\(\displaystyle \) \(=\) \(\displaystyle F_{k + 2} - 1 + F_{k + 1}\) Induction Hypothesis
\(\displaystyle \) \(=\) \(\displaystyle F_{k + 3} - 1\) Definition of Fibonacci Number

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


Therefore:

$\displaystyle \forall n \in \Z_{\ge 0}: \sum_{j \mathop = 0}^n F_j = F_{n + 2} - 1$

$\blacksquare$


Also presented as

This can also be seen presented as:

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

which is seen to be equivalent to the result given, as $F_0 = 0$.


Sources