binomial coefficient (3n chose n) is divisible by 3

voyage200

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Nov 7, 2006
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I have to prove that the binomial coefficint (3n) is divisible by 3 for all positive
__________________________________( n )
integers n.

where:
(3n) = 3n!/(n! (3n-n!)=3/(2n)!
(n )
but I don't know how to prove its divisible by 3. Please help.
 
(3nn)\displaystyle \Large {3n \choose n} is divisible by 3?

=(3n)!(3nn)!n!=(3n)!(2n)!(n)!=(3n)(3n1)!(3n1(n1))!(n1)!(n)=(3n)(n)(3n1n1)=3(3n1n1)\displaystyle =\Large \frac{(3n)!}{(3n-n)!n!}=\frac{(3n)!}{(2n)!(n)!}=\frac{(3n)(3n-1)!}{(3n-1-(n-1) )!(n-1)!(n)}=\frac{(3n)}{(n)} {3n-1 \choose n-1}=3 {3n-1 \choose n-1}

All that is left is to notice that (3n1n1)\displaystyle {3n-1 \choose n-1} is always an integer.
 
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