Sum of Sequence of Squares of Fibonacci Numbers

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Theorem

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


Then:

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

That is:

${F_1}^2 + {F_2}^2 + {F_3}^2 + \cdots + {F_n}^2 = F_n F_{n + 1}$


Proof

Proof by induction:

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

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


Basis for the Induction

$\map P 1$ is the case ${F_1}^2 = 1 = F_3 - 1$, which holds from the definition of Fibonacci numbers.

\(\displaystyle \sum_{j \mathop = 1}^1 {F_j}^2\) \(=\) \(\displaystyle {F_1}^2\)
\(\displaystyle \) \(=\) \(\displaystyle 1 \times 1\)
\(\displaystyle \) \(=\) \(\displaystyle F_1 \times F_2\)

demonstrating that $\map P 1$ holds.

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:

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


Then we need to show:

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


Induction Step

This is our induction step:

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

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


Therefore:

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

$\blacksquare$


Sources