Re: [中學] 99省二區1-3
※ 引述《ythung (費瑪連珠)》之銘言:
: 設數列{a(n)}滿足:a(1)=a(2)=1及
: a(n+2)={[a(n+1)]^2+a(n+1)+1}/a(n), n=1,2,3,...
: 試證:數列{a(n)}中任意相鄰兩項都是互質的整數。
遞迴數列很適合賺p幣 XD
a(n+2)/a(n+1) ~ a(n+1)/a(n), 可以猜到 a(n+2)=k*a(n+1)+ ...
而前幾項為 1,1,3,13,51,..
易見其滿足 a(n+2)=5a(n+1)-a(n)-1.
假設 a(k+2)=5a(k+1)-a(k)-1 對於 k=1,2,..,n-1 成立
則 a(n+2)=[a(n+1)^2+a(n+1)+1]/a(n)
=[(5a(n)-a(n-1)-1)^2 + (5a(n)-a(n-1)-1) +1]/a(n)
=25a(n)-10a(n-1)-5 + [a(n-1)^2+a(n-1)+1]/a(n)
Note that [a(n-1)^2+a(n-1)+1]/a(n) = a(n-2),
so a(n+2)=25a(n)-10a(n-1)-5 +a(n-2)
=5(5a(n)-a(n-1)-1)-(5a(n-1)-a(n-2)-1)-1
=5a(n+1)-a(n)-1.
Therefore, {a(n)} are all integers.
Also, since a(n)|a(n+1)^2+a(n+1)+1, gcd(a(n),a(n+1))=1.
--
※ 發信站: 批踢踢實業坊(ptt.cc)
◆ From: 131.215.6.212
推
04/04 17:04, , 1F
04/04 17:04, 1F
推
04/04 17:22, , 2F
04/04 17:22, 2F
推
04/04 19:33, , 3F
04/04 19:33, 3F
→
04/04 19:34, , 4F
04/04 19:34, 4F
→
04/04 19:39, , 5F
04/04 19:39, 5F
推
04/04 19:47, , 6F
04/04 19:47, 6F
→
04/04 19:48, , 7F
04/04 19:48, 7F
推
04/04 20:14, , 8F
04/04 20:14, 8F
→
04/04 20:16, , 9F
04/04 20:16, 9F
推
04/04 20:24, , 10F
04/04 20:24, 10F
→
04/04 20:25, , 11F
04/04 20:25, 11F
推
04/05 02:20, , 12F
04/05 02:20, 12F
→
04/05 02:21, , 13F
04/05 02:21, 13F
→
04/05 03:40, , 14F
04/05 03:40, 14F
→
04/05 03:41, , 15F
04/05 03:41, 15F
→
04/05 03:42, , 16F
04/05 03:42, 16F
推
04/05 08:34, , 17F
04/05 08:34, 17F
推
04/05 09:32, , 18F
04/05 09:32, 18F
→
04/05 09:34, , 19F
04/05 09:34, 19F
推
04/05 09:40, , 20F
04/05 09:40, 20F
→
04/05 09:40, , 21F
04/05 09:40, 21F
→
04/05 09:41, , 22F
04/05 09:41, 22F
討論串 (同標題文章)