# Binomial Theorem/Factorial proof

• Mar 17th 2009, 05:17 PM
ssj4Gogeta
Binomial Theorem/Factorial proof
Hi

Here's the question:
There exists a positive integer m for which the coefficient of the middle term in the expansion of (1+x)^(2m) is odd. True or False?

Our teacher solved it and I remember the answer is 'false', but I can't seem to remember how he did that. So basically we have to prove that (2m)!/(m! m!) can never be odd.

Sorry if the question is too elementary for this forum, or if this isn't the right place to post it.
(Btw, I'm new here. (Hi) And I'm 18. Sorry, I couldn't figure out how to properly use latex code. Is there a guide for it?)
• Mar 18th 2009, 03:01 PM
awkward
Quote:

Originally Posted by ssj4Gogeta
Hi

Here's the question:
There exists a positive integer m for which the coefficient of the middle term in the expansion of (1+x)^(2m) is odd. True or False?

Our teacher solved it and I remember the answer is 'false', but I can't seem to remember how he did that. So basically we have to prove that (2m)!/(m! m!) can never be odd.

Sorry if the question is too elementary for this forum, or if this isn't the right place to post it.
(Btw, I'm new here. (Hi) And I'm 18. Sorry, I couldn't figure out how to properly use latex code. Is there a guide for it?)

Hi,

For LaTex, see
http://www.mathhelpforum.com/math-help/latex-help/.

Here are two proofs.

First Proof. Notation: $\binom{n}{m} = \frac{n!}{m! (n-m)!}$ is a binomial coefficient, the number of ways to choose n objects taken m at a time.

$\binom{2m}{m} = \frac{(2m)!}{m! \; m!} = \frac{(2m) (2m-1) \cdots (m+1)}{m (m-1) \cdots (1)}$
$=2 \; \frac{(2m-1) (2m-2) \cdots (m+1)}{(m-1)(m-2) \cdots (1)}$
$=2 \; \frac{(2m-1) (2m-2) \cdots (m+1)}{(m-1)(m-2) \cdots (1)} \;\frac{m!}{m!}$
$= 2 \; \frac{(2m-1)!}{(m-1)! \; m!}$
$= 2 \; \binom{2m-1}{m-1}$

which is an even integer.

Second Proof: $\binom{2m}{m}$ is the number of combinations of 2m objects taken m at a time. If we have selected m objects, there are m objects left over (un-selected); the unselected objects form another combination of m objects taken from the original 2m. So we can pair every combination of size m uniquely with another (and different) combination of size m. Therefore the total number of combinations is even.
• Mar 19th 2009, 06:16 AM
ssj4Gogeta
Thanks. (Happy) Once I'm over with the exams I'll try to participate in the forum as much as I can.