更新1:
怎麼導出來的?
Σ j=1 到 n Σ [ (j-1)(n-j) ] = 1/6 (n^3-n3^2+8n)?
回答 (1)
2017-02-06 8:15 am
✔ 最佳答案
Σ j=1 to n_[ (j－1)(n－j) ]=Σ j=1 to n_(nj－j^2－n+j)
=－Σ j=1 to n_j^2+(n+1)Σ j=1 to n_j－nΣ j=1 to n_1
=－n(n+1)(2n+1)/6+(n+1)*n(n+1)/2－n^2
=(n/12)*[－2(n+1)(2n+1)+6(n+1)^2－12n]
=(n/12)*[－2(2n^2+3n+1)+6(n^2+2n+1)－12n]
=(n/12)*[(－4n^2－6n－2)+(6n^2+12n+6)－12n]
=(n/12)(2n^2－6n+4)
=(n^3－3n^2+2n)/6
收錄日期: 2021-04-30 22:10:43
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20170205235844AAegmgD