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Unformatted text preview: MATH 290 – WRITING ASSIGNMENT 3 – SOLUTIONS 1. Define the Fibonacci sequence { f 1 , f 2 , f 3 , ... } by f 1 = f 2 = 1 and for n > 2, f n = f n 1 + f n 2 . Use induction to prove that for all natural numbers n , f n = α n β n α β , where α = 1 + √ 5 2 and β = 1 √ 5 2 . (Hint: α and β are the solutions to the equation x 2 x 1 = 0.) Solution: The proof will proceed by induction. Assume that n = 1. Then f n = 1 and α 1 β 1 α β = 1. Hence the result holds for n = 1. We also need to prove the result for n = 2 since the recursion formula for f n only holds for n > 2. So assume n = 2. Then f 2 = 1 and α 2 β 2 α β = α + β = 1 + √ 5 2 + 1 √ 5 2 = 1 . Hence the result holds for n = 2. Let n ∈ N , n > 2 and assume that the result holds for all k ∈ { 1 , 2 , ..., n } . We must show that the result holds for n + 1. By the induction hypothesis and the definition of the Fibonacci numbers, f n +1 = f n + f n 1 = α n β n α β + α n 1 β n 1 α β = α n + α n 1 ( β n + β n 1 ) α β ....
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This note was uploaded on 03/05/2011 for the course MATH 290 taught by Professor Alligood,k during the Fall '08 term at George Mason.
 Fall '08
 Alligood,K
 Advanced Math, Natural Numbers

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