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標題:

mathematical induction

發問:

1. let α,β be roots of x^2-2x-1=0, where α>β.for any positive integers n , let U(n)=(1/8^0.5)(α^n-β^n)V(n)=(1/8^0.5)(α^n+β^n)a) show thhat U(n+2)=2U(n+1)+U(n)V(n+2)=2V(n+1)+V(n)bi)find U(1) and U(2)ii)suppose U(n) and U(n+1) are integers, deduce... 顯示更多 1. let α,β be roots of x^2-2x-1=0, where α>β.for any positive integers n , let U(n)=(1/8^0.5)(α^n-β^n) V(n)=(1/8^0.5)(α^n+β^n) a) show thhat U(n+2)=2U(n+1)+U(n) V(n+2)=2V(n+1)+V(n) bi)find U(1) and U(2) ii)suppose U(n) and U(n+1) are integers, deduce that U(n+2)is also an integers iii) Is U(n) an integer for all positive integers n? Give reason c)Is V(n) an integers for all positive integers n? Give reason 更新: (1/8^0.5) =1/(8^0.5) 更新 2: the ans says U(1)=1 U(2)=2

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(a) α+β=2,αβ=-1 (α^n+2-β^n+2) =(α+β)(α^n+1-β^n+1)+(αβ^n+1-βα^n+1) =2(α^n+1-β^n+1)+αβ(β^n-α^n) =2(α^n+1-β^n+1)-(β^n-α^n) U(n+2)=2U(n+1)+U(n) On the other hand (α^n+2+β^n+2) =(α+β)(α^n+1+β^n+1)-(αβ^n+1+βα^n+1) =2(α^n+1+β^n+1)-αβ(β^n+α^n) =2(α^n+1+β^n+1)+(β^n+α^n) V(n+2)=2V(n+1)+V(n) (b)(i) U(1)=(1/8^0.5)(α-β)=(3/4)^(0.5) U(2)=(1/8^0.5)(α-β)(α+β)=2(3/4)^(0.5) (ii) Using U(n+2)=2U(n+1)+U(n) (iii) No, U(1) is not an integer (c) No,V(1) is not an integer 2009-07-08 16:20:41 補充： Sorry b(i) is U(1)=1 and U(2)=2 (iii) is Yes by the result of part (ii)

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