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How would I prove the following statement:

$\forall n \in \mathbb{N}$,

$$\sum_{k=1}^{n} F_{k}^{2} = F_{n}F_{n+1}$$

where $F_{k}$ denotes the $k^{th}$ Fibonacci number.


I understand that I'd have to use mathematical induction to prove this, but how would I formally write out a proof for this?

I'm also a little confused about the logic/algebra behind Fibonacci, once I begin to solve this.

Thank you!

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closed as off-topic by Namaste, Daniel W. Farlow, Shailesh, Alex Provost, JonMark Perry Jul 12 '17 at 4:47

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    $\begingroup$ Verify it is true for $n=1$, then assume $$\sum_{k=1}^{n-1} F_{k}^{2} = F_{n-1}F_{n}$$ and add $F_n^2$ to both sides. $\endgroup$ – N74 Jul 11 '17 at 20:13
  • $\begingroup$ might not really be what you're after, but here's a nice visual proof. $\endgroup$ – Dando18 Jul 11 '17 at 20:22
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Note I'm using the definition of Fibonacci Numbers where: \begin{align*} F_1 & = 1 \\ F_2 & = 1 \end{align*} I'm considering $F_0$ to be "undefined" because it makes the sum on the LHS act a little weird. This is fine - we're using induction to prove it for $n\geq 1$, and can prove it ourselves for $n = 0$ potentially (or leave it be, as I am).

Base Case:

We need to verify it for $n = 1$. This just says that: $$\sum_{k = 1}^1 F_k^2 = F_1 F_2\implies 1^2 = 1\times 1$$ This is clearly true.

Inductive Hypothesis:

Assume that: $$\sum_{k = 1}^{n-1} F_k^2 = F_{n-1} F_{n}$$ Add $F_n^2$ to both sides: $$F_n^2 + \sum_{k =1}^{n-1}F_k^2 = F_{n-1}F_n + F_n^2$$ Note that we can rewrite the left hand side as $\sum_{k = 1}^n F_k^2$, and the right hand side as: $$F_{n-1}F_n + F_n^2 = F_n(F_{n-1}+F_n) = F_nF_{n+1}$$ where the last equality is from how the Fibonacci sequence is defined.

So, we have that the equality holding for the $n-1$th case implies it holds for the $n$th case, and we've thus finished the induction.

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Fibonacci is defined by

$$F_1=F_2=1$$ $$F_{n}=F_{n-1}+F_{n-2}$$

For $n=1$ the statement is true:

$$F_1^2=1^2=1\cdot 1=F_1\cdot F_2$$

Assume it is true for $n=m$:

$$\sum_{k=1}^{m}F_k^2=F_m F_{m+1}$$

This implies that it is true for $n=m+1$ : $$\sum_{k=1}^{m+1}F_k^2=\sum_{k=1}^{m}F_k^2+F_{m+1}^2=F_mF_{m+1}+F_{m+1}^2$$

$$=(F_{m+2}-F_{m+1})F_{m+1}+F_{m+1}^2=F_{m+1}F_{m+2}.$$

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Hint: $\;F_{n+1}=F_n+F_{n-1}\,$ by the definition of Fibonacci numbers. Multiplying by $\,F_n\,$ gives:

$$F_{n+1} F_n = F_n^2 + F_{n-1} F_n \quad\iff\quad F_n^2 = F_{n+1} F_n - F_{n} F_{n-1}$$

Summing the latter relation up from $2$ to $n$ and telescoping:

$$\require{cancel} \begin{align} \sum_{k=2}^{n} F_{k}^{2} &= (\cancel{\color{red}{F_3 F_2}} - F_2 F_1) + (\bcancel{\color{blue}{F_4 F_3}} - \cancel{\color{red}{F_3 F_2}}) + (\cancel{F_5F_4}-\bcancel{\color{blue}{F_4F_3}})+ \cdots + (F_{n+1} F_n - \cancel{F_{n} F_{n-1}}) \\ &= - 1 + F_{n+1} F_n \end{align} $$

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