Math MI

現在只做 n=k+1 的情況
(k+1)[(k+4)+1][(k+8)+1]
=[k(k+4) + k+k+4+1][(k+8)+1]
=[k(k+4) + 2k+k+5][(k+8)+1]
=k(k+4)(k+8) + k(k+4) + (2k+5)(k+8) + 2k+5
=k(k+4)(k+8) + k^2 + 4k +2k^2+16k+ 5k +40 +2k +5
=k(k+4)(k+8) + 3k^2 + 27k +45
=k(k+4)(k+8) + 3(k^2 + 9k + 15)
which is divisible by 3

Let P(n) be the proposition 'n(n+4)(n+8) is divisible by 3 for all positive integers n'.
When n=1,
L.H.S.=1(1+4)(1+8)=5x9=45, which is divisible by 3.
∴P(1) is true.
Assume that P(k) is true for k is a positive integer, i.e.k(k+4)(k+8)=3K, where K is also a positive integer.
When n=k+1,
L.H.S.=(k+1)(n+5)(n+9)
=n^3+15n^2+59n+45
n(n+4)(n+8)=n^3+12n^2+32n
∴n^3+15n^2+59n+45
=(n^3+12n^2+32n)+(3n^2+27n+45)
=3K+3(n^2+9n+45)
=3(K+n^2+9n+45)
Since k and K are both positive integer, 3(K+n^2+9n+45) is also a positive integer.
Also, since 3(K+n^2+9n+45) is a multiple of 3, it is also divisible by 3.
∴P(k+1) is also true.
By M.I., P(n) is true for all positive integers n.

2012-08-15 08:13:06 補充：
唔乘哂出黎應該唔得，因為要乘晒出黎至可以做因式分解。