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Re: [Axiom-developer] sum(binomial(t+i,i),i=0..k)
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From: |
Martin Rubey |
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Subject: |
Re: [Axiom-developer] sum(binomial(t+i,i),i=0..k) |
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Date: |
Tue, 12 Jan 2010 06:51:21 +0100 |
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User-agent: |
Gnus/5.11 (Gnus v5.11) Emacs/22.3 (gnu/linux) |
Michael Becker <address@hidden> writes:
> Hi,
>
>
>
> can someone explain the following answer of axiom?
>
>
>
> (12) -> sum(binomial(t+i,i),i=0..k)
>
> t + k t - 1
> (t + k + 1)( ) - t( )
> k - 1
> (12) -----------------------------
> t + 1
> Type: Expression(Integer)
What do want to know? It's entirely correct. The second binomial should
be interpreted as always zero, that's all.
Martin