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 \(k\):

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

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## Comments

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TopNewestStep 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.

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

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Not so interesting problem ,than other problems

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