# binomial theorem help

• Sep 22nd 2009, 04:38 PM
ezong
binomial theorem help
How do you show that (2n)!/[(n!)(2^(n+1))] is not an integer? Any hints or tips are appreciated.
• Sep 22nd 2009, 04:57 PM
Taluivren
Hi!

by induction prove that $\frac{(2n)!}{n!2^{n+1}}=\frac{k}{2}$ for some odd $k$:

$n=1$ is clear, $k=1$.

Supposing $\frac{(2n)!}{n!2^{n+1}}=\frac{k}{2}$ for some odd $k$ you get

$\frac{(2(n+1))!}{(n+1)!2^{n+2}} = \frac{(2n)!}{n!2^{n+1}} \cdot \frac{(2n+1)(2n+2)}{2(n+1)}= \frac{(2n)!}{n!2^{n+1}} \cdot (2n+1) = \frac{k(2n+1)}{2}$

since product of two odd numbers is odd, the inductive step is completed.
• Sep 24th 2009, 09:08 AM
Renji Rodrigo
Other solution
we can use the identity

$\prod^{n}_{k=1}(2k-1)=\frac{(2n)!}{2^{n}(n!)}$

dividing by 2 in both sides we have that $\frac{(2n)!}{n!2^{n+1}}$ is not an integer , because the first term is a product of odd numbers
• Sep 24th 2009, 02:00 PM
Soroban
Hello, ezong!

This can be done head-on . . .

Quote:

Show that $\frac{(2n)!}{n!\,2^{n+1}}$ is not an integer.

We have: . $(2n)! \;=\;1\cdot2\cdot3\cdot4\cdots 2n$

. . . . . . $= \;\bigg[ 1\cdot3\cdot5\cdots(2n-1)\bigg]\,\bigg[2\cdot4\cdot6\cdots2n\bigg]$

. . . . . . $= \;\bigg[1\cdot3\cdot5\cdot(2n-1)\bigg]\,2^n\bigg[1\cdot2\cdot3\cdots n\bigg]$

. . . . . . $= \;\bigg[1\cdot3\cdot5\cdots(2n-1)\bigg]\cdot2^n\cdot n!$

Then: . $\frac{(2n)!}{n!\,2^{n+1}} \;=\;\frac{\bigg[1\cdot3\cdot5\cdots(2n-1)\bigg]\cdot 2^n\cdot {\color{red}\rlap{//}}n!}{{\color{red}\rlap{//}}n!\,2^{n+1}} \;=\;\frac{1\cdot3\cdot5\cdots(2n-1)}{2}$

The numerator is the product of odd integers.
. . Hence, it is odd . . . of the form $2k-1$

Therefore: . $\frac{2k-1}{2}\:=\:k - \tfrac{1}{2}$ is not an integer.

• Oct 8th 2009, 02:30 PM
ezong
Thanks for the help!
Btw, how do you get the numbers to look so neat?
• Oct 9th 2009, 04:25 AM
Taluivren
Quote:

Originally Posted by ezong
Thanks for the help!
Btw, how do you get the numbers to look so neat?

Hello ezong, you can learn some basic LaTex typing here: http://www.mathhelpforum.com/math-he...-tutorial.html
and consult in LaTex Help subforum http://www.mathhelpforum.com/math-help/latex-help/