# need help with math simplification

Discussion in 'Homework Help' started by s3b4k, Mar 11, 2010.

1. ### s3b4k Thread Starter Member

Feb 15, 2010
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Jul 7, 2009
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Doesn't look like an equality to me -- plug in n = 1 and k = 0 (if you don't like those numbers, try n = 2 and k = 1).

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3. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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$\frac{(n-k)![(n-k+1)(n-k+2)-k(n-k+1)+k^2]}{k!(n-2k+2)!}$

is the closest I can come. Haven't checked it carefully.

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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Nope that didn't work either ...

Maybe ...

$\frac{(n-k)![(n-k+1)(n-k+2)-k(n-2k+1)-k^2]}{k!(n-2k+2)!}$

What's the point of the problem anyway?

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5. ### s3b4k Thread Starter Member

Feb 15, 2010
38
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because i needed to proove this identity and it was not equaling for me, just wanted to see if it would work for anyone else to see if I was making a math error