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 Post subject: Expanding a factorial (k is a constant)
PostPosted: Tue, 31 Jul 2012 03:09:49 UTC 
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FACTORIALS. How do you expand the below factorial, where k is a constant?

((k+8)(n+1))!


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 Post subject: Re: Expanding a factorial (k is a constant)
PostPosted: Tue, 31 Jul 2012 23:16:38 UTC 
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Joined: Mon, 19 May 2003 19:55:19 UTC
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Hello, A-R-Q!

We can do a little simplifying . . .


Quote:
How do you expand this factorial, where k is a constant? . [(k+8)(n+1)]!

We have: .[(k + 8)(n+1)]!


. . . . . =\;[(k+8)(n+1)]\,[(k+8)(n)]\,[(k+8)(n-1)]\,\cdots

. . . . . . . . . . [(k+8)(3)]\,[(k+8)(2)]\,[(k+8)(1)]


. . . . . =\;(k+8)^{n+1}\,[(n+1)(n)(n-1)\,\cdots\,3\cdot2\cdot1]


. . . . . =\;(k+8)^{n+1}(n+1)!



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 Post subject: Re: Expanding a factorial (k is a constant)
PostPosted: Tue, 31 Jul 2012 23:46:43 UTC 
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Joined: Fri, 27 Jul 2007 10:17:26 UTC
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Soroban wrote:
Hello, A-R-Q!

We can do a little simplifying . . .


Quote:
How do you expand this factorial, where k is a constant? . [(k+8)(n+1)]!

We have: .[(k + 8)(n+1)]!


. . . . . =\;[(k+8)(n+1)]\,[(k+8)(n)]\,[(k+8)(n-1)]\,\cdots

. . . . . . . . . . [(k+8)(3)]\,[(k+8)(2)]\,[(k+8)(1)]


. . . . . =\;(k+8)^{n+1}\,[(n+1)(n)(n-1)\,\cdots\,3\cdot2\cdot1]


. . . . . =\;(k+8)^{n+1}(n+1)!



Would it not be the following?

[(k+8)(n+1)][(k+8)(n+1) - 1][(k+8)(n+1) - 2]...

(k+8) would be a factor of the first term in [ ]'s of the factorial, but not the subsequent terms,
since (k+8)(n+1) - 1 \neq (k+8)n, etc.

Edit (outermeasure): Disable smilies.


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 Post subject: Re: Expanding a factorial (k is a constant)
PostPosted: Wed, 1 Aug 2012 14:10:44 UTC 
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Joined: Mon, 19 May 2003 19:55:19 UTC
Posts: 7949
Location: Lexington, MA
Hello, alstat!

Absolutely right! . . . *blush*

Everyone, please ignore my first post.



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