用數學歸納法 , 證明an = 3n+1, n=1,2,3,...
更新1:
a_k+1^2 - 3a_k+1 - (3k+4)(3k+1) = 0 [a_k+1 - (3k+4)][a_k+1 + (3k+1)] = 0 myisland8132也作出了貢獻 , 感謝
數學歸納法跪求詳解
2010-12-25 6:24 am
設{an} 的(上標)無限 (下標)n=1 為一正數列,使得a1+..+an = (an^2 + 3an - 4)/6 , n=1,2,3,...。
用數學歸納法 , 證明an = 3n+1, n=1,2,3,...
用數學歸納法 , 證明an = 3n+1, n=1,2,3,...
回答 (2)
2010-12-25 7:12 am
2010-12-25 7:27 am
a_1+....a_k+1=(a_k+1^2 + 3a_k+1 - 4)/6
3Σk+Σ1+a_k+1=(a_k+1^2 + 3a_k+1 - 4)/6
3k(k+1)/2+k+a_k+1=(a_k+1^2 + 3a_k+1 - 4)/6
9k(k+1)+6k+6a_k+1=(a_k+1^2 + 3a_k+1 - 4)
a_k+1^2 - 3a_k+1 - (9k^2+15k+4) = 0
a_k+1^2 - 3a_k+1 - (3k+4)(3k+1) = 0
[a_k+1 - (3k+4)][a_k+1 + (3k+1)] = 0
=> a_k+1=3k+4
3Σk+Σ1+a_k+1=(a_k+1^2 + 3a_k+1 - 4)/6
3k(k+1)/2+k+a_k+1=(a_k+1^2 + 3a_k+1 - 4)/6
9k(k+1)+6k+6a_k+1=(a_k+1^2 + 3a_k+1 - 4)
a_k+1^2 - 3a_k+1 - (9k^2+15k+4) = 0
a_k+1^2 - 3a_k+1 - (3k+4)(3k+1) = 0
[a_k+1 - (3k+4)][a_k+1 + (3k+1)] = 0
=> a_k+1=3k+4
收錄日期: 2021-04-23 23:24:23
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20101224000051KK01371