1. ## binomial theorem help

How do you show that (2n)!/[(n!)(2^(n+1))] is not an integer? Any hints or tips are appreciated.

2. 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.

3. 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

4. Hello, ezong!

This can be done head-on . . .

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.

5. Thanks for the help!
Btw, how do you get the numbers to look so neat?

6. 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/