Expanding a factorial (k is a constant)

A-R-Q · **Joined:** Sat, 20 Nov 2010 21:17:02 UTC **Posts:** 432

FACTORIALS. How do you expand the below factorial, where k is a constant?

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

Soroban · **Posted:** Tue, 31 Jul 2012 23:16:38 UTC

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)]!

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

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

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

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

alstat · **Posted:** Tue, 31 Jul 2012 23:46:43 UTC

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]...

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.

Soroban · **Posted:** Wed, 1 Aug 2012 14:10:44 UTC

Hello, alstat!

Absolutely right! . . . *blush*

Everyone, please ignore my first post.

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Expanding a factorial (k is a constant)

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