anonymous
  • anonymous
is (2n)! the same as 2n(n!)?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
No, because, if they were, dividing them would yield 1. Here, \[\frac{(2n)!}{2n(n!)}=\frac{2n(2n-1)(2n-2)...(n+1)n!}{2n \times n!}\]\[=(2n-1)(2n-2)...(n+1)\neq1\]
nowhereman
  • nowhereman
at least for n > 1 ;-)
anonymous
  • anonymous
Yes, I sort of assumed.

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anonymous
  • anonymous
how does the (2n)! turn into the (n+1)n! at the end there?
nowhereman
  • nowhereman
thats because he reduces n! and the first factor that remains is (n+1)n!
anonymous
  • anonymous
does that mean (2n)! = 2n(n+1)n! ?
anonymous
  • anonymous
thats not right at all, sorry.
anonymous
  • anonymous
i am trying to make it look like n! = (n+1)n!, you know?
anonymous
  • anonymous
\[(n+1)!=(n+1)n!\]
nowhereman
  • nowhereman
yes indeed \[(2n)! = 2n(2n-1)(2n-2)\cdots(2n-(n-1))n!\]
anonymous
  • anonymous
oh i see how that happened now. thanks for spelling it out for me :)
anonymous
  • anonymous
Fan us, then, bugaloo! ;)

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