Re: [中學] 利用數學歸納法

看板Math作者s00459 (沉靜)時間13年前 (2010/12/31 02:32)推噓1(1推 0噓 6→)

留言7則, 2人參與討論串3/4 (看更多)

※ 引述《billiechick (比利小雞)》之銘言： : 利用數學歸納法 : 對任意正整數n,證明n^5-n為30的倍數 : 拜託各位了! 當 n=1 , 1^5 - 1 = 0 成立設 n=k 成立，則 k^5 - k = k(k-1)(k+1)(k^2 + 1) = 30m 當 n=k+1 時 (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)有四個連續數，故必有5、2、3的因數，則為30的倍數令 5k(k+1)(k-1)(k+2)=30p 15k(k+1)有兩個連續整數，必有2的因數，再乘上15也必為30的倍數令15k(k+1)=30q 故(k+1)^5-(k+1) = 30m+30p+30q=30(m+p+q)為30的倍數成立所以對任意正整數n，n^5-n必為30的倍數成立 -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 114.40.134.181

推

Holocaust123

12/31 02:39, , 1^F

12/31 02:39, 1^F

→

theoculus

12/31 03:16, , 2^F

12/31 03:16, 2^F

→

theoculus

12/31 03:16, , 3^F

12/31 03:16, 3^F