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The Lucas numbers satisfy the recurrence relation Ln=Ln-1+ Ln-2, and the initial conditions are Lo = 2 and 4₁ = 1. Click and drag statements to find an explicit formula for the Lucas numbers. L₁ = α₂ -a for some constants a, and a₂. The recurrence relation L₁= L-1 + L₂-2 has characteristic equation 1+√√5 2+7-1=0. Its roots are >= - L-a(1+√³)* + a₂ (1-√³) 1+√√5 1-√√5 L₁ = 2 for some constants a, and a₂. 1+√ 4- (1+³)+(1-³). L₁=2 and L₁=1-a₁=1 and a₂ = 1. L₂ = 4-(+)-(-). L₁= 2 and L₁ = 1 → a₁ = 1 and ₂ = 1. L = The recurrence relation L=L+L2 has characteristic equation 1+√5 2 2-1=0. Its roots are >=

The Lucas numbers satisfy the recurrence relation Ln=Ln-1+ Ln-2, and the initial conditions are Lo = 2 and 4₁ = 1. Click and drag statements to find an explicit formula for the Lucas numbers. L₁ = α₂ -a for some constants a, and a₂. The recurrence relation L₁= L-1 + L₂-2 has characteristic equation 1+√√5 2+7-1=0. Its roots are >= - L-a(1+√³)* + a₂ (1-√³) 1+√√5 1-√√5 L₁ = 2 for some constants a, and a₂. 1+√ 4- (1+³)+(1-³). L₁=2 and L₁=1-a₁=1 and a₂ = 1. L₂ = 4-(+)-(-). L₁= 2 and L₁ = 1 → a₁ = 1 and ₂ = 1. L = The recurrence relation L=L+L2 has characteristic equation 1+√5 2 2-1=0. Its roots are >=

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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The Lucas numbers satisfy the recurrence relation Ln=Ln-1+ Ln-2, and the initial conditions are Lo = 2 and L₁ = 1. Click and drag statements to find an explicit formula for the Lucas numbers. L = a (¹+√³)² - a₂ (1-√³) +√√5 for some constants α₁ and ₂. The recurrence relation L₂= L-1 + L-2 has characteristic equation 1+√√5 2 ²+1=0. Its roots are r = 1-√√√5 L₂ = a (¹+√³)² + a₂ (¹ = √³)* 2 for some constants a₁ and ₂. 1+ 1.4. - (¹ + √/³)* + (¹-1/³). Lo=2 and L₁=1 → a= 1 and ₂ = 1. L₁ = 4- - (¹ + ³) - (¹ -- ). Lo=2 and L₁ = 1 → α₁ = 1 and α₂ = 1. L₁ = The recurrence relation L₂=Ln-1 + Ln-2 has characteristic equation 1+√√5 2 72-1=0. Its roots are r =
Transcribed Image Text:The Lucas numbers satisfy the recurrence relation Ln=Ln-1+ Ln-2, and the initial conditions are Lo = 2 and L₁ = 1. Click and drag statements to find an explicit formula for the Lucas numbers. L = a (¹+√³)² - a₂ (1-√³) +√√5 for some constants α₁ and ₂. The recurrence relation L₂= L-1 + L-2 has characteristic equation 1+√√5 2 ²+1=0. Its roots are r = 1-√√√5 L₂ = a (¹+√³)² + a₂ (¹ = √³)* 2 for some constants a₁ and ₂. 1+ 1.4. - (¹ + √/³)* + (¹-1/³). Lo=2 and L₁=1 → a= 1 and ₂ = 1. L₁ = 4- - (¹ + ³) - (¹ -- ). Lo=2 and L₁ = 1 → α₁ = 1 and α₂ = 1. L₁ = The recurrence relation L₂=Ln-1 + Ln-2 has characteristic equation 1+√√5 2 72-1=0. Its roots are r =
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