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Click and drag expressions to show that Ln= fn−1+fn+1 for n=2,3,..., where fn is the nth Fibonacci number. If P(2) AA P(k), then Lk+1 = = Basis Step:P(1) is true because L₁-1 and f+f2-0 +1 -1. P(2) is true because L₂-L₁ | Lo-2|1-3 and 2-1f2+1-1 2-3. Inductive Step. Assume P() is true for all / with 2 ≤jsk for some arbitrary k, 2 sks, that is, assume L;=fj+f; forj=2,...,k fk + fx + 2 (fx-2+ fx+1)+(fx + fx+1) Lk + Lk-1 Let P(n) be the proposition that Ln=fn− 1 + fn+ 1. (fx-1+fx+1)+(fx-2 + fx)

Click and drag expressions to show that Ln= fn−1+fn+1 for n=2,3,..., where fn is the nth Fibonacci number. If P(2) AA P(k), then Lk+1 = = Basis Step:P(1) is true because L₁-1 and f+f2-0 +1 -1. P(2) is true because L₂-L₁ | Lo-2|1-3 and 2-1f2+1-1 2-3. Inductive Step. Assume P() is true for all / with 2 ≤jsk for some arbitrary k, 2 sks, that is, assume L;=fj+f; forj=2,...,k fk + fx + 2 (fx-2+ fx+1)+(fx + fx+1) Lk + Lk-1 Let P(n) be the proposition that Ln=fn− 1 + fn+ 1. (fx-1+fx+1)+(fx-2 + fx)

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter1: Fundamental Concepts Of Algebra
Section1.2: Exponents And Radicals
Problem 92E
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Click and drag expressions to show that n= fn − 1 + fn + 1 for n = 2, 3, ..., where fɲ is the nth Fibonacci number. If P(2)^^ P(k), then Lk+1 = = = = Basis Step:P(1) is true because L₁ - 1 and fy+f2-0 +1 -1. P(2) is true because L₂ - L₁ | Lo - 2 | 1 − 3 and ƒ2 −1 |ƒ2 +1-1 | 2-3. Inductive Step. Assume P(j) is true for all j with 2 ≤ j≤k for some arbitrary k, 2 ≤ k≤n, that is, assume L;= fj-1 + fj+1 for j = 2,...,.k. fix + fix + 2 (fx-2+ fx+1)+(fx + fx+1) Lk + Lk-1 Let P(n) be the proposition that Ln=fn − 1 + fn + 1. (fk-1 + fx+1)+(fix-2 + fix)
Transcribed Image Text:Click and drag expressions to show that n= fn − 1 + fn + 1 for n = 2, 3, ..., where fɲ is the nth Fibonacci number. If P(2)^^ P(k), then Lk+1 = = = = Basis Step:P(1) is true because L₁ - 1 and fy+f2-0 +1 -1. P(2) is true because L₂ - L₁ | Lo - 2 | 1 − 3 and ƒ2 −1 |ƒ2 +1-1 | 2-3. Inductive Step. Assume P(j) is true for all j with 2 ≤ j≤k for some arbitrary k, 2 ≤ k≤n, that is, assume L;= fj-1 + fj+1 for j = 2,...,.k. fix + fix + 2 (fx-2+ fx+1)+(fx + fx+1) Lk + Lk-1 Let P(n) be the proposition that Ln=fn − 1 + fn + 1. (fk-1 + fx+1)+(fix-2 + fix)
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