Re: recognizing integer numbers

*To*: mathgroup at smc.vnet.net*Subject*: [mg34040] Re: [mg33924] recognizing integer numbers*From*: BobHanlon at aol.com*Date*: Sat, 27 Apr 2002 00:57:12 -0400 (EDT)*Sender*: owner-wri-mathgroup at wolfram.com

In a message dated 4/23/02 8:27:20 AM, manuel_ballester at yahoo.com writes: >I need some help with the following. I am trying mathematica to solve >some sums for me. The thing is that I don't know how to tell >mathematica that certain numbers are non-negative integers and some >answers that I would expect as (for example) n! are given as >Gamma[n-1] and so on, sometimes Hypergeometric functions are also >involved. Example: > >Sum[Binomial[2*m+1,k]*( a^k*b^(2*m+1-k)+ b^k*a^(2*m+1-k) ),{k,0,m}] > >if m is a natural number then the answer to this is simply >(a+b)^(2*m+1) > >but since I don't know how to say this to mathematica it gives me an >answer with gamma and hypergeometric functions. I already tried >Simplify and FullSimplify. > s1 = FullSimplify[Sum[Binomial[2*m+1,k]* (a^k*b^(2*m+1-k)+b^k*a^(2*m+1-k)), {k,0,m}], Element[m, Integers]]; The same sum in reverse order is s2 = FullSimplify[Sum[(Binomial[2*m+1,k]* (a^k*b^(2*m+1-k)+b^k*a^(2*m+1-k))) /. k -> (m-k),{k,0,m}], Element[m, Integers]]; The result is then Simplify[(s1+s2)/2] (a + b)^(2*m + 1) Bob Hanlon Chantilly, VA USA