利用數學歸納法,証明對於所以正整數n , 下列命題成立
9.1/2*5 + 1/5*8 + 1/8*11 + ... + 1/(3n-1)(3n+2) = n/6n + 4
急,我有d關於A Maths o既問題想問吓大家 (MI)
2008-11-04 3:24 am
回答 (1)
2008-11-04 4:34 am
✔ 最佳答案
For n = k+1.1/2*5 + 1/5*8 + ......+ 1/[3(k+1) - 1][3(k+1) + 2]
= k/(6k +4) + 1/(3k+2)(3k+ 5)
= [k(3k+5) + 2]/2(3k+2)(3k+5)
= [3k^2 + 5k +2]/2(3k+2)(3k+5)
= (3k + 2)(k + 1)/2(3k+2)(3k+5)
= (k+1)/2(3k+5)
= (k+1)/(6k + 10)
= (k+1)/[6(k+1) + 4].
收錄日期: 2021-04-25 22:35:26
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20081103000051KK01520