V voyage200 New member Joined Nov 7, 2006 Messages 4 Nov 7, 2006 #1 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.
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.
M marcmtlca Junior Member Joined Oct 22, 2006 Messages 87 Nov 8, 2006 #2 \(\displaystyle \Large {3n \choose n}\) is divisible by 3? \(\displaystyle =\Large \frac{(3n)!}{(3n-n)!n!}=\frac{(3n)!}{(2n)!(n)!}=\frac{(3n)(3n-1)!}{(3n-1-(n-1) )!(n-1)!(n)}=\frac{(3n)}{(n)} {3n-1 \choose n-1}=3 {3n-1 \choose n-1}\) All that is left is to notice that \(\displaystyle {3n-1 \choose n-1}\) is always an integer.
\(\displaystyle \Large {3n \choose n}\) is divisible by 3? \(\displaystyle =\Large \frac{(3n)!}{(3n-n)!n!}=\frac{(3n)!}{(2n)!(n)!}=\frac{(3n)(3n-1)!}{(3n-1-(n-1) )!(n-1)!(n)}=\frac{(3n)}{(n)} {3n-1 \choose n-1}=3 {3n-1 \choose n-1}\) All that is left is to notice that \(\displaystyle {3n-1 \choose n-1}\) is always an integer.
