DISCRETE MATH: Binomial Theorem proof (using Corollary 2)

[VinnyCee]

Homework Statement

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

Homework Equations

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

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

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.

Thanks for your help in advance.

 

[mjsd]

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

 

[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?

 

[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 ?

 

[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?

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

 

[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?

 

[mjsd]

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