中四Ａ.Ｍａｔｈ　（１）

let S(n) be the statement
對任意正整數n,(2n^3+n)可被3整除.

when n=1
2+1=3 is divisible by 3

So, S(1) is true

when n=k
assume S(k) is true
that is (2k^3+k)=3M (where M is a constant)

when n=k+1
2(k+1)^3+k+1
=2(k^3+3k^2+3k+1)+k+1
=2k^3+k+6k^2+6k+3
=3M+6k^2+6k+3
=3(M+2k^2+2k+1)

which is divisible by 3
So, when n=k+1, S(k+1) is true

by MI, for all positive values of n, S(n) is true

2006-12-13 20:24:15 補充：
set p( n )= 2n^3 np (1)=2 1=3所以p(1)是正確假設 p( k ) 是正確所以 2k^3 k=3m當n= k 1=2( k 1)^3 k 12( k^3 3k^2 3k 1 ) k 1=2k^3 6k^2 4k 22k^3 6k^2 4k 3=3m-k 6k^2 4k 3=3m 6k^2 3k 3=3(m 2k^2 k 1)所以可以比3除

set p( n )= 2n^3+n
p (1)=2+1=3
所以p(1)是正確
假設 p( k ) 是正確
當n= k+1
=2( k +1)^3+k+1
2( k^3+3k^2+3k+1 ) +k +1
=2k^3+6k^2+4k+2
2k^3+6k^2+4k+3
=2(3m-k)+6k^2+4k+3
=3(2m)+6k^2+3k+3
=3(2m+2k^2+k+1)
所以可以比3除

2006-12-13 17:51:53 補充：
set p( n )= 2n^3 np (1)=2 1=3所以p(1)是正確假設 p( k ) 是正確所以 2k^3 k=3m當n= k 1=2( k 1)^3 k 12( k^3 3k^2 3k 1 ) k 1=2k^3 6k^2 4k 22k^3 6k^2 4k 3=3m-k 6k^2 4k 3=3m 6k^2 3k 3=3(m 2k^2 k 1)所以可以比3除