A Proof of Combinatorial Identity

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