# Can $\sum_{i=0}^n (k^i)/i!$ be streamlined?

Can this expression be streamlined?

$ \sum_{i=0}^n (k^i)/i! $

EDIT : This expression I have obtained as the variety of feasible means to select n things or much less from k various boundless things (you can select as several as you can from any kind of type) ...

I think it must amount to the variety of feasible means to select n things from k+1 various boundless things, where the added+1 is a dummy object, picking among this type suggests allowing a void in the last option (i.e. picking n - 1 not n) ...

so this must amount to :

$ (k+1)^n/n! $

Why that is wrong??

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Betamoo 2019-12-02 02:50:55

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Answers: 1

In feedback to the modified inquiry, it is wrong. If we simply allow n = 2, $1+k+\frac{k^2}{2}\neq \frac{(k+1)^2}{2}$. The distinction is $\frac{1}{2}$

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Ross Millikan 2019-12-03 04:17:57

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