附加數,歸納法題!!

let f(n)= 2n^3+n
for n=0
f(0)=0=0x3
設 f(k) 能被３整除，因此我們得著
2k^3+k=3Q-----(1)
for n=k+1
f(k+1)=2(k+1)^3+(k+1)
=2(k^3+3k^2+3k+1)+(k+1)
=2k^3+k +6k^2+6k+3
=2k^3+k+3(2k^2+2k+1)
=3Q+3(2k^2+2k+1)
=3(Q+2k^2+2k+1)
therefore f(k+1) is also divisible by 3
hence by M.I. f(n) is divisible by 3 for all n=positive integers.
現在讓我們思想n=negative integers.
since f(-n)=2(-n)^3+(-n)=-(2n^3+n)=-f(n)
just because f(n) is divisible by 3, then f(-n) is also divisible by 3
hence, we have f(n) is divisible by 3 for all n=integers.


2008-01-13 23:55:05 補充：
對唔住！英文有少許問題，應該是：hence by M.I. f(n) is divisible by 3 for all positive integers n.hence, we have f(n) is divisible by 3 for all integers n.

2008-01-13 23:56:43 補充：
注意：這條沒有說明只證明n是正整數時，所以我們也要證明n是負整數也對。正只要n是整數，那(2n^3 n)可被3整除

let s(n) be the statement,
when n = 1, 2 x 1 + 1 = 3 which is divisible by 3.
s(1) is true.
assume s(k) is true, 2k^3 + k = 3m where m is a integer.
when n = k + 1,
= 2(k+1)^3 + k + 1
= 2(k^3 + 3k^2 + 3k + 1) + k + 1
= 2k^3 + 6k^2 + 6k + 2 + k + 1
= 3m + 6k^2 + 6k + 3
= 3(m + 2k^2 + 2k + 1)
By the principle of M.I. (2n^3+n)可被3整除.

2008-01-13 17:14:10 補充：
= 2k^3 + 6k^2 + 6k + 2 + k + 1= 3m + 6k^2 + 6k + 3用左assumption^^