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 $k$ :

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

#Algebra

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`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
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Comments

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.

Curtis Clement - 6 years, 1 month ago

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

Victor Paes Plinio - 6 years, 1 month ago

Not so interesting problem ,than other problems

Kutumbaka Jaswanth - 6 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$