中四附加數學

Let P(n)be the proposition：‘n(n+1)(n+2)(n+3) is divisible by 12
When n = 1，
1(1+1)(1+2)(1+3) = 24 = 12(2)
which is divisible by 12
∴P(1) is true
Assume that P(n) is true，
i.e.k(k+1)(k+2)(k+3) = 12m，where m is an integer
when n = k+1，
(k+1)[(k+1)+1][(k+1)+2][(k+1)+3]
(k+1)(k+2)(k+3)(k+4)
= k(k+1)(k+2)(k+3) + 4(k+1)(k+2)(k+3)
= 12m+ 4(k+1)(k+2)(k+3)
Note that for any 3 consecutive integers, one of them must be divisible by 3.
Thus (k + 1)(k + 2)(k + 3) is divisible by 3. [Actually you can prove it by M.I.]
Write (k + 1)(k + 2)(k + 3) = 3N, where N is an integer. Then
(k + 1)(k + 2)(k + 3)(k + 4)
= 12M + 4(3N) = 12(M + N), where M + N is an integer.
Thus (k + 1)(k + 2)(k + 3)(k + 4) is divisible by 12.
Thus it is true for n = k + 1.
By the principle of mathematical induction, it is true for any positive integers n.

n=(n+2)*5/3