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1.3+2.4+3.5+...+n﹝n+2﹞=n﹝n+1﹞﹝an+b﹞/6,則數對﹝a,b﹞=?

2005-12-02 11:14:43 · 3 個解答 · 發問者 Anonymous in 科學 數學

為什麼我都看不懂呢~= =

2005-12-02 16:01:09 · update #1

3 個解答

原式=
[summation 2--> n+1] (n-1)(n+1)
= [summation 2--> n+1] (n^2) - [summation 2--> n+1] (1^2)
=[ (n+1)(n+2)(2n+3) / 6 - 1^2 ] - (n)
展開後得
(2n^3 + 9n^2 + 7n) / 6
即 n ( n + 1 )( n + 7 ) 故a=1 b=7

2005-12-02 12:09:11 · answer #1 · answered by 仲宇 黃 1 · 0 0

因 n(n+2)=n^2+2n
故 原式 1.3+2.4+3.5+...+n﹝n+2﹞
= [1^2+2^2+3^2+..........+n^2] + 2[1+2+3+............+n]
= [n(n+1)(2n+1)] /6 + 2[n(n+1)/2]
= n(n+1)[(2n+1) /6+1]
= n(n+1)(2n+7) /6
故 a=2 b=7

2005-12-02 18:30:52 · answer #2 · answered by fumi 6 · 0 0

設原式會=Σ(從1~n)k(k+1)
=Σk^2+2Σk
=n(n+1)(2n+1)/6+2n(n+1)/2
=n(n+1)(2n+7)
→ a=2 , b=7

2005-12-02 15:18:36 · answer #3 · answered by 小渝 1 · 0 0

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