中四符數(歸納法) 聖誕假期工課

Let P(n) be the proposition:
(√3+1)^(2n+1)-(√3-1)^(2n+1) is divisible by 2^(n+1)
when n=1
LHS
=(√3+1)^3-(√3-1)^3
=20
=4*5
So P(1) is true
when n=2
LHS
=(√3+1)^5-(√3-1)^5
=152
=8*19
So P(2) is true
Assume that P(k) and P(k+1) is true
(√3+1)^(2k+1)-(√3-1)^(2k+1)=2^(k+1)M
(√3+1)^(2k+3)-(√3-1)^(2k+3)=2^(k+1)N
when n=k+2
LHS
=(√3+1)^(2k+5)-(√3-1)^(2k+5)
=[(√3+1)^(2k+3)-(√3-1)^(2k+3)]][(√3+1)^2+(√3-1)^2]-[(√3+1)^(2k+3)(√3-1)^2-(√3-1)^(2k+3)(√3+1)^2]
=8*2^(k+2)N-[(√3+1)^(2k+1)(√3+1)^2(√3-1)^2-(√3-1)^(2k+1)(√3-1)^2(√3+1)^2]
=8*2^(k+2)N-4[(√3+1)^(2k+1)-(√3-1)^(2k+1)]
=8*2^(k+2)N-4*2^(k+1)M
=2^(k+3)[4N-M]
So P(k+2) is true
By the principle of mathematical induction,P(n) is true for all positive integers n.

2007-12-11 15:34:02 補充：
17行應是(√3+1)^(2k+3)-(√3-1)^(2k+3)=2^(k+2)N