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【趣味数学题】垛积术(高阶等差级数)

2021-10-03 09:52 作者:AoiSTZ23  | 我要投稿

 郑涛(Tao Steven Zheng)著

【问题】

宋元时期(公元 960年 - 1368年)的数学家研究了与贾宪三角对角线有关的有限级数(finite seies)。以下列表给出了朱世杰《算学启蒙》(1299)和《四元玉鉴》(1303)中的级数。

茭草垛
1%2B2%2B3%2B4%2B...%2Bn%20%3D%20%5Cfrac%7B1%7D%7B2!%7D%20n(n%2B1)

落一形垛
1%2B3%2B6%2B10%2B...%2B%5Cfrac%7B1%7D%7B2!%7Dn(n%2B1)%20%3D%20%5Cfrac%7B1%7D%7B3!%7Dn(n%2B1)(n%2B2)%20

撒星形垛
1%2B4%2B10%2B20%2B...%2B%5Cfrac%7B1%7D%7B3!%7Dn(n%2B1)(n%2B2)%20%3D%20%5Cfrac%7B1%7D%7B4!%7Dn(n%2B1)(n%2B2)(n%2B3)

撒星更落一形垛
1%2B5%2B15%2B35%2B...%2B%20%5Cfrac%7B1%7D%7B4!%7Dn(n%2B1)(n%2B2)(n%2B3)%3D%5Cfrac%7B1%7D%7B5!%7Dn(n%2B1)(n%2B2)(n%2B3)(n%2B4)

这些有限级数叫高阶等差级数(arithmetic series of higher order),遵循一般公式

%5Csum_%7Bi%3D1%7D%5E%7Bn%7D%5Cfrac%7B1%7D%7Bp!%7Di(i%2B1)(i%2B2)...(i%2Bp-1)%20%3D%20%5Cfrac%7B1%7D%7B(p%2B1)!%7Dn(n%2B1)(n%2B2)...(n%2Bp)%20

也可以写成

%20%5Csum_%7Bi%3D1%7D%5E%7Bn%7D%20%5Cbinom%7Bi%2Bp-1%7D%7Bp%7D%20%3D%20%5Cbinom%7Bn%2Bp%7D%7Bp%2B1%7D

证明这个恒等式。


【题解】

%5Csum_%7Bi%3D1%7D%5E%7Bn%7D%20%5Cbinom%7Bi%2Bp-1%7D%7Bp%7D%20%3D%20%5Cbinom%7Bp%7D%7Bp%7D%20%2B%20%5Cbinom%7Bp%2B1%7D%7Bp%7D%20%2B%20%5Cbinom%7Bp%2B2%7D%7Bp%7D%20%2B%20...%20%2B%20%5Cbinom%7Bn%2Bp-1%7D%7Bp%7D

注意 %20%5Cbinom%7Bp%7D%7Bp%7D%20%3D%20%5Cbinom%7Bp%2B1%7D%7Bp%2B1%7D,所以

%5Csum_%7Bi%3D1%7D%5E%7Bn%7D%20%5Cbinom%7Bi%2Bp-1%7D%7Bp%7D%20%3D%20%5Cleft%5B%5Cbinom%7Bp%2B1%7D%7Bp%2B1%7D%20%2B%20%5Cbinom%7Bp%2B1%7D%7Bp%7D%20%5Cright%5D%20%2B%20%5Cbinom%7Bp%2B2%7D%7Bp%7D%20%2B%20...%20%2B%20%5Cbinom%7Bn%2Bp-1%7D%7Bp%7D%20

使用二项式恒等式 %20%5Cbinom%7Bn-1%7D%7Bk%7D%20%2B%20%5Cbinom%7Bn-1%7D%7Bk-1%7D%20%3D%20%5Cbinom%7Bn%7D%7Bk%7D,得

%5Csum_%7Bi%3D1%7D%5E%7Bn%7D%20%5Cbinom%7Bi%2Bp-1%7D%7Bp%7D%20%3D%20%5Cbinom%7Bp%2B2%7D%7Bp%2B1%7D%20%2B%20%5Cbinom%7Bp%2B2%7D%7Bp%7D%20%2B%20...%20%2B%20%5Cbinom%7Bn%2Bp-1%7D%7Bp%7D%20

遵循这个步骤模式

%5Csum_%7Bi%3D1%7D%5E%7Bn%7D%20%5Cbinom%7Bi%2Bp-1%7D%7Bp%7D%20%3D%20%5Cleft%5B%5Cbinom%7Bp%2B2%7D%7Bp%2B1%7D%20%2B%20%5Cbinom%7Bp%2B2%7D%7Bp%7D%20%5Cright%5D%20%2B%20%5Cbinom%7Bp%2B3%7D%7Bp%7D%20%2B%20...%20%2B%20%5Cbinom%7Bn%2Bp-1%7D%7Bp%7D%20

直到最后两项

%5Csum_%7Bi%3D1%7D%5E%7Bn%7D%20%5Cbinom%7Bi%2Bp-1%7D%7Bp%7D%20%3D%20%5Cleft%5B%5Cbinom%7Bn%2Bp-1%7D%7Bp%2B1%7D%20%2B%20%5Cbinom%7Bn%2Bp-1%7D%7Bp%7D%20%5Cright%5D%20

因此,

%20%5Csum_%7Bi%3D1%7D%5E%7Bn%7D%20%5Cbinom%7Bi%2Bp-1%7D%7Bp%7D%20%3D%20%5Cbinom%7Bn%2Bp%7D%7Bp%2B1%7D



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