diff --git a/chap1/1.2/1.2.6/1.2.6.tex b/chap1/1.2/1.2.6/1.2.6.tex index 5d71456..5385421 100644 --- a/chap1/1.2/1.2.6/1.2.6.tex +++ b/chap1/1.2/1.2.6/1.2.6.tex @@ -1179,4 +1179,20 @@ Thus if $n m$, we have $a_{n,m} = n a_{n-1,m}$. So finally \[ a_{n,m} = [ n \ge m ] \frac{n!}{m!}.\] + +\newpar{61} Note $S(m,n,l) = \sum_k(-1)^k {l+m \choose l+k}{m+n + \choose m+k}{n+l \choose n+k}$. Using \textbf{exercise 31}, we have +\begin{eqnarray*} + S(m,n,l) &=& \sum_k(-1)^k {l+m \choose l+k}{m+n \choose m+k}{n+l + \choose l-k} \\ + &=& \sum_k (-1)^k {l+m \choose l+k} \sum_i {k+l \choose i} {m-k + \choose l-k-i}{m+n+i \choose m+l} \\ + &=& \sum_k \sum_i \frac{(m+n+i)!}{(n+i-l)! i! (m-l+i)!} \times + \frac{(-1)^k}{(k+l-i)!(l-k-i)!} \\ + &=& \sum_i \frac{(m+n+i)!}{(n+i-l)!i!(m-l+i)!(2l-2i)!} \sum_k (-1)^k + {2(l-i) \choose k+l-i} \\ + &=& \sum_i \frac{(m+n+i)!}{(n+i-l)!i!(m-l+i)!(2l-2i)!} \times + (-1)^{l+i} \delta_{l,i} \\ + &=& \frac{(m+n+l)!}{m!n!l!} +\end{eqnarray*} \end{document}