作者s00459 (沉静)
看板Math
标题Re: [中学] 利用数学归纳法
时间Fri Dec 31 02:32:14 2010
※ 引述《billiechick (比利小鸡)》之铭言：
: 利用数学归纳法
: 对任意正整数n,证明n^5-n为30的倍数
: 拜托各位了!
当 n=1 , 1^5 - 1 = 0 成立
设 n=k 成立，则 k^5 - k = k(k-1)(k+1)(k^2 + 1) = 30m
当 n=k+1 时
(k+1)^5 - (k+1) = k(k+1)(k+2)(k^2+2k+2)
= k(k+1)(k-1+3)(k^2+2k+2)
= k(k-1)(k+1)(k^2+2k+2)+3k(k+1)(k^2+2k+2)
= k(k-1)(k+1)(k^2+1+2k+1)+3k(k+1)(k^2+2k+2)
= k(k-1)(k+1)(k^2+1)+3k(k+1)(k^2+2k+2)+k(k-1)(k+1)(2k+1)
= 30m+k(k+1)[3(k^2+2k+2)+(k-1)(2k+1)]
= 30m+k(k+1)[3k^2+6k+6+2k^2-k-1]
= 30m+k(k+1)(5k^2+5k+5)
= 30m+5k(k+1)(k^2+k+1)
= 30m+5k(k+1)(k^2+k-2+3)
= 30m+5k(k+1)(k^2+k-2)+15k(k+1)
= 30m+5k(k+1)(k-1)(k+2)+15k(k+1)
由於 5k(k+1)(k-1)(k+2)有四个连续数，故必有5、2、3的因数，则为30的倍数
令 5k(k+1)(k-1)(k+2)=30p
15k(k+1)有两个连续整数，必有2的因数，再乘上15也必为30的倍数
令15k(k+1)=30q
故(k+1)^5-(k+1) = 30m+30p+30q=30(m+p+q)为30的倍数 成立
所以对任意正整数n，n^5-n必为30的倍数成立
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※ 发信站: 批踢踢实业坊(ptt.cc)
◆ From: 114.40.134.181
1F：推 Holocaust123:赞! 12/31 02:39
2F：→ theoculus :(k+1)^5-(k+1) = k^5-k+5(k+2k^2+2k^3+k^4) 12/31 03:16
3F：→ theoculus : = 30m +5k(k+1)(k^2+k+1) 12/31 03:16
4F：→ theoculus : = 30m +5k(k+1){k^2+4k+4-3(k+1)} 12/31 03:17
5F：→ theoculus : = 30m +5k(k+1){(k+2)^2 - 3(k+1)} 12/31 03:18
6F：→ theoculus : = 30m +5k(k+1)(k+2)^2 - 15k(k+1)^2 12/31 03:18
7F：→ theoculus :这样会简短一点 12/31 03:18