$\displaystyle

k^k - {k\choose 1}\left( {k - 1} \right)^k + {k\choose 2}\left( {k - 2} \right)^k - {k\choose 3}\left( {k - 3} \right)^k + \ldots + \left( { - 1} \right)^k {k\choose k}\left( {k - k} \right)^k=?

$

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- Apr 5th 2008, 07:33 AMjames_bondBinomial coefficients
$\displaystyle

k^k - {k\choose 1}\left( {k - 1} \right)^k + {k\choose 2}\left( {k - 2} \right)^k - {k\choose 3}\left( {k - 3} \right)^k + \ldots + \left( { - 1} \right)^k {k\choose k}\left( {k - k} \right)^k=?

$ - Apr 5th 2008, 07:38 AMMathstud28If you are asking
what that series is equal to i think its $\displaystyle \sum_{n=0}^{k}{k\choose n}(k-n)$

- Apr 5th 2008, 07:47 AMTrevorP
This what I got in terms of a summation.

$\displaystyle

k^k + \sum_{n=1}^{k}{k\choose n}(k-n)^k(-1)^n

$ - Apr 5th 2008, 07:48 AMMathstud28Yeah
hes right I did not see the alternating signs

- Apr 5th 2008, 11:11 AMjames_bond
I suppose that the answer should be given in a closed form. I can see that it can be written shorter using the sum symbol ($\displaystyle \sum_{n=0}^{k}{k\choose n}(k-n)^k(-1)^n$)...

- Apr 5th 2008, 11:21 AMPlato