奧數題(10點)

👨🏻‍🏫 學術台數學

嘉浩

2009-05-06 1:58 am

1x2+2x3+3x4+4x5+......9x10=
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)

回答 (3)

Victor

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

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[匿名]

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

參考: me

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myisland8132

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

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