數學歸納法(2)

Part 1
Let P(n) be the proposition:
bn is an even integer for all positive integers n
when n=1
b1=[(1+√3)+(1-√3)]=2
So P(1) is true
Assume that P(k) is true
bk=(1+√3)^k+(1-√3)^k is an even integer
when n=k+1
bk+1
=(1+√3)^(k+1)+(1-√3)^(k+1)
=(1+√3)(1+√3)^k+(1-√3)(1-√3)^k
=(1+√3)^k+(1-√3)^k+√3[(1+√3)^k-(1-√3)^k]
=2M+√3(2K) [where M and K is an integer]
(Using binomial theorem (1+√3)^k-(1-√3)^k is an even integer)
=2(M+√3K)
which is an even integer
P(k+1) is true

By the principle of mathematical induction,P(n) is true for all positive integers n.

Part 2
Let P(n) be the proposition:
an is an integer for all positive integers n
when n=1
a1=1/(2√3)[(1+√3)-(1-√3)]=1
So P(1) is true
Assume that P(k) is true
ak=1/(2√3)[(1+√3)^k-(1-√3)^k] is an integer
when n=k+1
ak+1
=1/(2√3)[(1+√3)^(k+1)-(1-√3)^(k+1)]
=1/(2√3)[(1+√3)(1+√3)^k-(1-√3)(1-√3)^k]
=1/(2√3)[(1+√3)^k-(1-√3)^k+√3[(1+√3)^k+(1-√3)^k] ]
=M+K
where
M=1/(2√3)[(1+√3)^k-(1-√3)^k]
K=1/(2√3)[√3[(1+√3)^k+(1-√3)^k]]
is an integer by Part 1 and the assumption of ak
So P(k+1) is true

By the principle of mathematical induction,P(n) is true for all positive integers n.