【分享】Σnk次方公式及推導(倒)過程


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原網頁:http://tw.knowledge.yahoo.com/question/question?qid=1006122906714

由二項式定理可導出Σnk 的通式:Σnk =(1/C1k+1).

[(n+1)k+1 -C2k+1 (Σnk- 1) -C3k+1 (Σnk-2) - …… -Ckk+1 (Σn) - (n+1)]

以下是推導過程:

1k+1 = ………… = 1

2k+1 = (1+1)k+1 = 1k+1 + C1k+1.1k + C2k+1.1k- 1 +……+ Ckk+1.1 1 + 1

3k+1 = (2+1)k+1 = 2k+1 + C1k+1.2k + C2k+1.2k- 1 +……+ Ckk+1.2 1 + 1

4k+1 = (3+1)k+1 = 3k+1 + C1k+1.3k + C2k+1.3k- 1 +……+ Ckk+1.3 1 + 1

…… (n+1)k+1 = nk+1 + C1k+1.nk + C2k+1.nk- 1 +……+ Ckk+1.n1 + 1

上列等式相加,消去共同項(紅色):

(n+1)k+1 = C1k+1.(Σnk) + C2k+1.(Σnk- 1)+……+ Ckk+1.(Σn) + (n+1)

⇒Σnk =(1/C1k+1)[(n+1)k+1-C2k+1 (Σnk- 1) -……-Ckk+1 (Σn)-(n+1)]

舉些例子:

Σn1 =(1/C12)[(n+1) 2 - (n+1)]

= n(n+1)/2

Σn2 = (1/C13)[(n+1) 3 - C23 (Σn) - (n+1)]

=(1/3)[(n+1) 3 - 3.n(n+1)/2 - (n+1)]

= n(n+1)( 2n+1)/6

Σn3 = (1/C14)[(n+1) 4 - C24 (Σn2) -C34 (Σn) - (n+1)]

=(1/4)[(n+1) 4 - 6(n(n+1)( 2n+1)/6) - 4(n(n+1)/2) - (n+1)]

= n2 (n+1) 2/4

Σn4 = (1/5)[(n+1) 5 - 10.(Σn3)- 10.(Σn2)- 5.(Σn)- (n+1)]

= (1/5)[(n+1)5 - 10( n2 (n+1) 2 /4)- 10(n(n+1)( 2n+1)/6)

- 5(n(n+1)/2)- (n+1) ] = ……

= n(n+1)(2n+1)(3n2 +3n-1)/30

Σn5 = (1/6)[(n+1) 6 - 15.(Σn4)- 20.(Σn3)- 15.(Σn2)- 6.(Σn)- (n+1)]= ……

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