呢種數應該點計
更新1:
唔明,可唔可以簡單d
更新2:
第三個比較簡單,請您加以解釋,(如果加以解釋我會出送分題,20點)
奧數題(10點)
回答 (3)
2009-05-06 2:31 am
✔ 最佳答案
1x2+2x3+3x4+4x5+......9x10= 9x10x11 /3 = 330
其實對於所有n, r,
都有
1 x 2 x 3 x ... x n + 2 x 3 x 4x ...x (n+1) + ...... + (r+1) x (r+2) x ...x (r+n)
= (r+1) x (r+2) x ... x (r+n) x (r+n+1) / (n+1)
如：
1x2x3x4 + 2x3x4x5 + 3x4x5x6 +......+ 10x11x12x13
=10x11x12x13x14 / 5
如果要證明這個property，可用數學歸納法
參考: 自己
2009-05-06 2:15 am
奧數題
1x2+2x3+3x4+4x5+......9x10
=Σi(i+1) [i from 1 to 9]
=Σ(i^2+i)
=9(9+1)(2*9+1)/6+9(9+1)/2
=330
1x2+2x3+3x4+4x5+......9x10
=Σi(i+1) [i from 1 to 9]
=Σ(i^2+i)
=9(9+1)(2*9+1)/6+9(9+1)/2
=330
參考: me
2009-05-06 2:02 am
1x2+2x3+3x4+4x5+......9x10
=Σi(i+1) [i from 1 to 9]
=Σ(i^2+i)
=9(9+1)(2*9+1)/6+9(9+1)/2
=330
=Σi(i+1) [i from 1 to 9]
=Σ(i^2+i)
=9(9+1)(2*9+1)/6+9(9+1)/2
=330
收錄日期: 2021-04-26 13:14:43
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090505000051KK01074