AMATHS的mi題目啊..急...10分!!

當N=k+1
(k+1)(k+2)(2k+3)
= k(k+1)(2k+3) + 2(k+1)(2k+3)
=k(k+1)(2k+1) + 2k(k+1) +2(k+1)(2k+3)
= 3 M + 2 ( k^2 + k + 2k^2 + 5k + 3 )
= 3 M + 6 (k^2 + 2k + 1 )
= 3 [ M + 2(k^2 + 2k +1)]

Let S(n) be the statement n(n+1)(2n+1) is divisible by 3 for all netural numbers n.
Let f(n) = n(n+1)(2n+1),
f(1) = (1)(2)(3)
= 6, which is divisible by 3.
所以 S(1) is true.

Assume S(k) is true,
i.e. f(k) = k(k+1)(2k+1) = 3m, where m is an interger.

f(k+1) = (k+1)(k+2)[2(k+1)+1]
= (k+1)(k+2)(2k+3)
= [ 3m/[(k)(2k+1)] ] (k+2)(2k+3)
= (6mk^2 + 21mk + 18) / (2k^2+k)
=(3) [(2mk^2 + 7mk + 6)/ (k^2+k) ]
因為m, k are intergers.
所以f(k+1) is divisible by 3
所以S(k+1) is true.
所以By the mathematical induction, S(n) is true for all natural intergers n.