Ta = (a^2 + 10(a!)!),其中a為任意正整數 (Ta唔係乘既關係,因為唔識打)
利用數學歸納法,證明對任意正整數
T1+T2+……Ta=a[(a+1)!] (T1唔係乘既關係,係T下一個小1)
[註:a!=a(a-1)(a-2)…3*2*1]
A MATH
2009-01-04 11:37 pm
回答 (3)
2009-01-04 11:47 pm
✔ 最佳答案
題目錯了吧!? 應該是T(a)= (a^2+1)*a! n=1時, T(1)=2*1!= 2 = 1*2! 成立
設n=k時, T(1)+...+T(k)=k*(k+1)!
n=k+1時
T(1)+...+T(k) +T(k+1)
=k*(k+1)! + [(k+1)^2 + 1]*(k+1)!
=(k+k^2+2k+1+1)(k+1)!
=(k+1)(k+2)*(k+1)!
=(k+1)*(k+2)! 成立
由數學歸納法原理,故得證
參考: me
2009-01-04 11:51 pm
http://hk.knowledge.yahoo.com/question/question?qid=7008123102111
煩惱即是菩提﹐這條幾有趣﹐但給人破解了
煩惱即是菩提﹐這條幾有趣﹐但給人破解了
2009-01-04 11:43 pm
LHS=T1=11=/=RHS
Is it real???
Is it real???
收錄日期: 2021-04-26 13:05:52
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090104000051KK01142