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# Fibonacci Question

The Fibonacci Sequence is given by the following rule:

$$\begin{cases} { F }_{ 1 }=1 \\ { F }_{ 2 }=1 \\ { F }_{ n }={ F }_{ n-2 }+{ F }_{ n-1 } \end{cases}$$

Proof that for $$k \ge 2$$ the following rule is valid for any value for $$k$$:

${ { F }_{ k } }^{ 2 }={ F }_{ k-1 }\cdot { F }_{ k+1 }+{ \left( -1 \right) }^{ k-1 }$

Note by Victor Paes Plinio
2 years, 10 months ago

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Step 1: For n = 2: LHS = 1^2 = 1 , RHS = (1 $$\times$$ 2) -1 = 1 so the formula is shown to be valid thus far. Step2: Assume that the formula is valid for all n = k Step 3: Now it's time for the induction step..... Firstly, rearrange the equation: $\ (F_k)^2 - F_{k+1}F_{k-1} = (-1)^{k-1}$ Now using $$\ F_{k-1} = F_{k+1} - F_{k}$$ $\ (F_k)^2 - F_{k+1}F_{k-1} = (F_k)^2 - F_{k+1} (F_{k+1} - F_{k}) = (F_k)^2 - (F_{k+1})^2 + F_{k}F_{k+1} = (-1)^{k-1}$ $F_k (F_k +F_{k+1}) - (F_{k+1})^2 = F_{k}F_{k+2} - (F_{k+1})^2 = (-1)^{k-1}$ Now let n = k+1 $\therefore\ (F_{n})^2 - F_{n-1}F_{n+1} = (-1)^{n-1}$ We have proven that the formula is true for n=2, and once we assumed that it was true for n = k then it was true for n = k+1. Hence, the formula must be true for $$\ k \geq\ 2$$ Q.E.D.

- 2 years, 10 months ago

Excellent! The induction method is a nice way to solve this type of problem!

- 2 years, 10 months ago

Not so interesting problem ,than other problems

- 2 years, 10 months ago