兩條公式一問(中學先學o既)

1.你係咪漏左第一項 1x2 ?

1x2 + 2x3 + 3x4 +..... +Nx(N+1)
=1(1 + 1) + 2(2+1) +3(3+1) + 4(4+1) + ... + N(N +1)
=1^2 + 2^2 + 3^2 + 4^2 + ... + N^2 + (1 + 2 + 3 + 4 + ... + N)
= N(N+1)(2N+1)/6 + N(N+1)/2
= N(N+1) x [(2N + 1)/6 + 1/2]
= N(N+1) x (2N + 1 + 3)/6
= N(N+1)(2N+4)/6
= N(N+1) x [ 2(N+2)/6]
= 1/3 x N(N+1)(N+2)

2. 同樣漏了第一項 1x2x3.
首先: Nx(N+1)x(N+2) = N^3 + 3N^2 +2N
所以: 1x2x3 + 2x3x4 + 3x4x5 +... + Nx(N+1)x(N+2)
= (1^3 + 2^3 + 3^3 + ... + N^3) + 3(1^2 + 2^2 + 3^2 + ... +N^2)
　+ 2(1 + 2 + 3 + ... + N)
= N^2(N+1)^2 / 4 + 3 [ N(N+1)(2N+1)/6 ] + 2[ N(N+1)/2 ]
= N^2(N+1)^2 / 4 + N(N+1)(2N+1)/2 + N(N+1)
= N(N+1) [ N(N+1) / 4 + (2N+1)/2 + 1 ]
= N(N+1) [ (N^2 + N + 4N + 2 + 4) / 4 ]
= 1/4 x N(N+1) (N^2 + 5N + 6)
= 1/4 x N(N+1)(N+2)(N+3)

2007-08-06 16:43:50 補充：
呢d 公式全部係由以下三條基本公式而來:一: 1 ＋ 2 ＋ 3 ＋ ... ＋ n = n(n＋1)/2二: 1^2 ＋ 2^2 ＋ 3^2 ＋ ... ＋ n^2 = n(n＋1)(2n＋1)/6三: 1^3 ＋ 2^3 ＋ 3^3 ＋ ... ＋ n^3 = (1 ＋ 2 ＋ 3 ＋ ... ＋ n)^2 　　= [n(n＋1)/2]^2 = n^2(n＋1)^2/4方法係將原本的數列改為由以上三組數列表示。一你都知點prove 啦! 三角形數二、三點prove 我都唔知

1x2+2x3+3x4+......Nx(N+1)
=1^2+1+2^2+2+3^3+3+....+N^2+N
=(1^2+2^2+3^2+.....+N^2)+(1+2+3+...+N)
=(1/6)(N)(N+1)(2N+1)+(1/2)(N)(N+1)
=(1/6)(N)(N+1)(2N+1)+(1/6)(3N)(N+1)
=(1/6)(N)(N+1)[2N+1+3)
=(1/3)(N)(N+1)(N+2)//

PS:1^2+2^2+...+N^2=(1/6)(N)(N+1)(2N+1)
1+2+3+.....+N=(1/2)(N)(N+1)

1x2x3+2x3x4+3x4x5+......N(N+1)(N+2)
=1^3+2x1^2+3x1+2^3+2x2^2+3x2+....+N^3+2N^2+3N
=[N(N+1)/2]^2+2(1/6)(N)(N+1)(2N+1)+(3/2)(N)(N+1)
=N(N+1)[(1/4)(N)(N+1)+(1/3)(2N+1)+(3/2)]
=N(N+1){(1/12)[(3N)(N+1)+4(2N+1)+18]
=(1/12)(N)(N+1)[3N^2+3N+4N+4+18]
=(1/12)(N)(N+1)(3N^2+7N+22)
=(1/4)(N)(N+1)(N+2)(N+3)//