已知N是整數,求證(3N-1)(3N+2)是偶數
回答 (3)
2007-12-15 4:11 am
✔ 最佳答案
首先,要答案是一個偶數,(3N-1)或(3N+2)至少其中之一是偶數假如3N是一個偶數,(3N+2)一定是偶數,所以答案是偶數
假如3N是一個奇數,(3N-1)一定是偶數,所以答案也是偶數
2007-12-15 8:19 am
設p(N)為命題
p(N)=(3N-1)(3N+2)
when N=1
p(1)=2x5,是偶數
∴p(1)成立
假設p(K)成立,k≧1
p(K)=(3K-1)(3K+2)=2M,M為正整數
∴(3K-1)=2M/(3K+2)
when N=K+1
p(K+1)=[3(K+1)-1][3(K+1)+2]
=(3K+2)(3K+5)
=(3K+2)(3K-1+6)
=(3K+2) [2M/(3K+2) + 6]
=2M+ 6(3K+2)
=2[M+3(3K+2)] , 是偶數
∴p(K+1) 成立
根據數學歸納法原理,對於一切正整數N,命題p(N)成立
p(N)=(3N-1)(3N+2)
when N=1
p(1)=2x5,是偶數
∴p(1)成立
假設p(K)成立,k≧1
p(K)=(3K-1)(3K+2)=2M,M為正整數
∴(3K-1)=2M/(3K+2)
when N=K+1
p(K+1)=[3(K+1)-1][3(K+1)+2]
=(3K+2)(3K+5)
=(3K+2)(3K-1+6)
=(3K+2) [2M/(3K+2) + 6]
=2M+ 6(3K+2)
=2[M+3(3K+2)] , 是偶數
∴p(K+1) 成立
根據數學歸納法原理,對於一切正整數N,命題p(N)成立
參考: me, 中五 method
2007-12-15 4:33 am
(3N-1)(3N+2)
=3N(3N+2)-1(3N+2)
=9N^2+6N-3N-2
=8N^2+2N-2+N^2+N
=2(4N^2+N-1)+N(N+1)
since N(N+1) can be divided by 2 whenever N is odd or even,
so (3N-1)(3N+2) can be also divided by 2.
2007-12-14 20:38:19 補充：
IF N is odd-- N 1 is even....N(N 1) also evenIF N is even --) N(N 1) is even
=3N(3N+2)-1(3N+2)
=9N^2+6N-3N-2
=8N^2+2N-2+N^2+N
=2(4N^2+N-1)+N(N+1)
since N(N+1) can be divided by 2 whenever N is odd or even,
so (3N-1)(3N+2) can be also divided by 2.
2007-12-14 20:38:19 補充：
IF N is odd-- N 1 is even....N(N 1) also evenIF N is even --) N(N 1) is even
收錄日期: 2021-04-29 19:59:27
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20071214000051KK02896