# DISCRETE MATH: Binomial Theorem proof (using Corollary 2)

1. Mar 28, 2007

### VinnyCee

1. The problem statement, all variables and given/known data

Show that if n is a positive integer, then $$1\,=\,\binom{n}{0}\,<\,\binom{n}{1}\,<\,\cdots\,<\,\binom{n}{\lfloor\frac{n}{2}\rfloor}\,=\,\binom{n}{\lceil\frac{n}{2}\rceil}\,>\,\cdots\,>\binom{n}{n\,-\,1}\,>\,\,\binom{n}{n}\,=\,1$$

2. Relevant equations

I think this proof involves corollary 2 of the Binomial Theorem.

$$\sum_{k\,=\,0}^n\,(-1)^k\,\binom{n}{k}\,=\,0$$

3. The attempt at a solution

I have NO idea where to start a proof of this sort. This problem is labeled as 'easy', but I don't see the easiness yet. I know that an attempt at a solution is required before any assistance is rendered, but I have been staring at (and thinking about) this problem for two days now. I can't think of ANYTHING that is even remotely helping me to "show" this theorem.

I am not trying to have someone else do my homework or anything like that, I just need a pointer or two. If you don't believe me, look at my previous threads, all have detailed attempts and most were completed problems after help was given.

Last edited: Mar 28, 2007
2. Mar 28, 2007

### mjsd

have you tried using the definition
$$\binom{n}{k}=\frac{n!}{k!(n-k)!}$$
to help?

3. Mar 28, 2007

### VinnyCee

Yeah, I used that up until the ceiling and floor functions in the middle. I can show that 1 is equal to the next term and that the term after that is indeed larger than the term equal to one. But what do I do with the ceiling and floor functions that come after? How to show that they are larger?

4. Mar 28, 2007

### mjsd

what can you and what can't you prove? you can't prove for the cases k > n/2 or k < n/2 or just the middle terms ?

5. Mar 28, 2007

### VinnyCee

By definition, $\binom{n}{0}\,=\,1$. Now, $\binom{n}{1}\,=\,\frac{n!}{(n\,-\,1)!}$. And by definition of the factorial function $n!\,>\,(n\,-\,1)!$. Also, a divisor that is less than the dividend results in a quotient that is greater than one. This proves up to the point of $\binom{n}{1}$, and I think it can be extended up to the part where I am stuck. How do I work with the floor functions using the definition of binomial coefficient?

$$\binom{n}{\lfloor\frac{n}{2}\rfloor}$$

6. Mar 28, 2007

### VinnyCee

I guess I have to consider two cases of the center equality (the one using the ceiling and floor functions) in order to prove the whole inequality. There is a case of when $n$ is even and the case of when $n$ is odd. In the even case, the two terms are obviously equal.

$n$ = 4:

$$\binom{4}{\lfloor\frac{4}{2}\rfloor}\,=\,\binom{4}{2}\,=\,\frac{4!}{2!(4\,-\,2)!}\,=\,\frac{4!}{2!}\,=\,\frac{24}{2}\,=\,12$$

$$\binom{4}{\lceil\frac{4}{2}\rceil}\,=\,\binom{4}{2}\,=\,\frac{4!}{2!(4\,-\,2)!}\,=\,\frac{4!}{2!}\,=\,\frac{24}{2}\,=\,12$$

Okay, but how do I prove anything about the odd $n$-values?

$n$ = 5:

$$\binom{5}{\lfloor\frac{5}{2}\rfloor}\,=\,\binom{5}{2}\,=\,\frac{5!}{2!(5\,-\,2)!}\,=\,\frac{5!}{2!\,3!}\,=\,\frac{120}{12}\,=\,10$$

$$\binom{5}{\lceil\frac{5}{2}\rceil}\,=\,\binom{5}{3}\,=\,\frac{5!}{3!(5\,-\,3)!}\,=\,\frac{5!}{2!\,3!}\,=\,\frac{120}{12}\,=\,10$$

$n$ = 7:

$$\binom{7}{\lfloor\frac{7}{2}\rfloor}\,=\,\binom{7}{3}\,=\,\frac{7!}{3!(7\,-\,3)!}\,=\,\frac{7!}{3!\,4!}\,=\,\frac{5040}{144}\,=\,35$$

$$\binom{7}{\lceil\frac{7}{2}\rceil}\,=\,\binom{4}{4}\,=\,\frac{7!}{4!(7\,-\,4)!}\,=\,\frac{7!}{4!\,3!}\,=\,\frac{5040}{144}\,=\,35$$

I guess that proves that in either case the two are equal! Now how do I combine it all into a concise proof?

Last edited: Mar 28, 2007
7. Mar 28, 2007

### mjsd

do a case for even n and a case for odd n