Sum of Sequence of Squares of Fibonacci Numbers
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Theorem
Let $F_k$ be the $k$th Fibonacci number.
Then:
- $\forall n \ge 1: \ds \sum_{j \mathop = 1}^n {F_j}^2 = F_n F_{n + 1}$
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:
- $\ds \sum_{j \mathop = 1}^n {F_j}^2 = F_n F_{n + 1}$
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.
\(\ds \sum_{j \mathop = 1}^1 {F_j}^2\) | \(=\) | \(\ds {F_1}^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 1 \times 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds F_1 \times F_2\) |
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:
- $\ds \sum_{j \mathop = 1}^k {F_j}^2 = F_k F_{k + 1}$
Then we need to show:
- $\ds \sum_{j \mathop = 1}^{k + 1} {F_j}^2 = F_{k + 1} F_{k + 2}$
Induction Step
This is our induction step:
\(\ds \sum_{j \mathop = 1}^{k + 1} {F_j}^2\) | \(=\) | \(\ds \sum_{j \mathop = 1}^k {F_j}^2 + {F_{k + 1} }^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds F_k F_{k + 1} + {F_{k + 1} }^2\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {F_k + F_{k + 1} } F_{k + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds F_{k + 2} F_{k + 1}\) | Definition of Fibonacci Number |
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\forall n \ge 1: \ds \sum_{j \mathop = 1}^n {F_j}^2 = F_n F_{n + 1}$
$\blacksquare$
Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $5$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $5$