✔ 最佳答案
(a) Sn+2 = 3n+2[(1 + √2)n+2 + (1 - √2)n+2]
= (9)(3n)[(1 + √2)n(1 + √2)2 + (1 - √2)n(1 - √2)2]
= (9)(3n)[(1 + √2)n(1 + 2√2 + 2) + (1 - √2)n(1 - 2√2 + 2)]
= (9)(3n)[(1 + √2)n(2√2 + 2) + (1 + √2)n + (1 - √2)n(- 2√2 + 2) + (1 - √2)n]
= (9)(3n)[2(1 + √2)n(1 + √2) + 2(1 - √2)n(1 - √2) + (1 + √2)n + (1 - √2)n]
= (18)(3n)[(1 + √2)n(1 + √2) + (1 - √2)n(1 - √2)] + 9(3n)[(1 + √2)n + (1 - √2)n]
= (6)(3n+1)[(1 + √2)n(1 + √2) + (1 - √2)n(1 - √2)] + 9Sn
= 6Sn+1 + 9Sn
(b) Let P(n) be the proposition that Sn is divisible by 3n
P(1) is true since S1 = 3[(1 + √2) + (1 - √2)] = 6 is divisible by 3
P(2) is true since S2 = 3[(1 + √2)2 + (1 - √2)2] = 3(1 + 2√2 + 2 + 1 - 2√2 + 2) = 18 is divisible by 32
Let P(k) and P(k+1) be true, i.e. Sk = 3kN and Sk+1 = 3k+1M where M and N are integers
Sk+2 = 6Sk+1 + 9Sk
= 6(3k+1M) + 9(3kN)
= 2(3k+2)M + 3k+2N
= 3k+2(2M + N) so Sk+2 is divisible by 3k+2 => P(k+2) is true
So by the principle of MI, P(n) is true for all positive integer n.
2009-09-18 21:11:40 補充：
Please see below for a better view of the answer:
http://img19.imageshack.us/img19/9624/86065421.png
