mathematical induction

用戶9285

2009-07-09 12:02 am

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:

(1/8^0.5) =1/(8^0.5)

更新2:

the ans says U(1)=1 U(2)=2

回答 (1)

myisland8132

2009-07-09 12:18 am

✔ 最佳答案

(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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