From: peter.luschny@gmail.com   
      
   What about this approach:   
      
   Replace the last two occurrences of n by k in   
       sum(binomial(n-i-1,n-i), i=0..n);   
      
       sum(binomial(n-i-1,k-i), i=0..k);   
      
   Now this sum is (for 0 <= k < n)   
      
       -n*binomial(n-1, k)/(k-n)   
      
       a := (n,k) -> n*binomial(n-1, k)/(n-k);   
      
   for n from 0 to 5 do seq(a(n,k),k=0..n-1) od;   
      
              1   
             1, 2   
           1, 3, 3   
          1, 4, 6, 4   
       1, 5, 10, 10, 5   
      
   So how do you want this (quite famous) triangle to be   
   extended on the right hand side? OK, I agree.   
      
   The most simple trick will do: look at the limit.   
      
   for n from 0 to 5 do seq(limit(a(n,m),m=k),k=0..n) od;   
      
               0   
              1, 1   
            1, 2, 1   
           1, 3, 3, 1   
         1, 4, 6, 4, 1   
       1, 5, 10, 10, 5, 1   
      
   Fine. Only the case (0,0) is razor-ed here by Maple.   
   But in this case we have   
      
       sum(binomial(0-i-1,0-i), i=0..0) = 1   
      
   So there is no problem here and the general answer is:   
      
       sum(binomial(n-i-1,n-i), i=0..n) = 1   
      
   by extension by limit of sum(binomial(n-i-1,k-i),i=0..k).   
      
   Still I am worried that Mathematica insists on   
      
       sum(binomial(n-i-1,n-i), i=0..n) = 1.   
      
   P.S.:   
   Caveat: When I say 'Mathematica' I really mean 'Alpha'.   
   I just received a message of Hans Haverman who   
   alerts me that in some of these equations   
   "Alpha's result disagrees with Mathematica's."   
      
   --- SoupGate-Win32 v1.05   
    * Origin: you cannot sedate... all the things you hate (1:229/2)