在這篇論文中，我們主要是研究一個組合等式如下：∑_(i=0)^n▒∑_(j=0)^i▒〖C(n,i)C(n+1,j)=?〗
在解這個等式時，我們將不使用一般的計算方式：而採用了建構一個對射函數（bijective function）的方法，進而得到上面等式的解。
接著我們推廣此等式為∑_(i=0)^n▒∑_(j=0)^i▒〖C(n,i)C(n+m,j)=?〗時，我們仍將繼續利用此函數是一對一的特性，為此組合等式求得通解如下，來完成這篇論文。∑_(i=0)^n▒∑_(j=0)^i▒〖C(n,i)C(n+m,j)=2^(2n+m-1)-〗 ∑_(i=0)^n▒∑_(j=1)^(m-1)▒C(n,i)C(n+m-1,i+j) / In this paper, we will mainly study a combinatorial identity, as the following：∑_(i=0)^n▒∑_(j=0)^i▒〖C(n,i)C(n+1,j)=?〗. When solving this identity, we will not use common calculation. Instead, we will build a method of bijective function in order to obtain the solution to the above identity.
To finish this paper, we will continue to generalize this identity as ∑_(i=0)^n▒∑_(j=0)^i▒〖C(n,i)C(n+m,j)=?〗 Then we will be able to use 1-1 property of this function as to get the following solution to the combinatorial identity：∑_(i=0)^n▒∑_(j=0)^i▒〖C(n,i)C(n+m,j)=2^(2n+m-1)-〗 ∑_(i=0)^n▒∑_(j=1)^(m-1)▒C(n,i)C(n+m-1,i+j)
| Identifer | oai:union.ndltd.org:CHENGCHI/B2002002894 |
| Creators | 陳建霖, Chen, Chien-Lin |
| Publisher | 國立政治大學 |
| Source Sets | National Chengchi University Libraries |
| Language | 英文 |
| Detected Language | English |
| Type | text |
| Rights | Copyright © nccu library on behalf of the copyright holders |
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