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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
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