Proving an inequality concerning binomial coefficents

How do I prove:$$\left(\begin{array}{l}n \\ 0\end{array}\right)<\left(\begin{array}{l}n \\ 1\end{array}\right)<\cdots<\left(\begin{array}{c}n \\ \lfloor n / 2\rfloor\end{array}\right)=\left(\begin{array}{c}n \\ \lceil n / 2\rceil\end{array}\right)>\cdots>\left(\begin{array}{c}n \\ n-1\end{array}\right)>\left(\begin{array}{l}n \\ n\end{array}\right)$$ I tried using the definition of the binomial coeffient. First of all, I split it up into to parts:

$$(1)$$ Prove: $$\begin{pmatrix} n\\m \end{pmatrix}<\begin{pmatrix} n\\k \end{pmatrix}$$ for $$k>m\in \{1,2,...,\lfloor n/2 \rfloor -1\}$$

$$(2)$$ Prove: $$\begin{pmatrix} n\\m \end{pmatrix}<\begin{pmatrix} n\\k \end{pmatrix}$$ for $$k

• Hint: Consider $\frac{n\choose k}{n\choose{k+1}}$. Oct 31, 2021 at 14:42

$$\frac{n+1}{k+1}{n\choose k}={n+1\choose k+1}={n\choose k+1}+{n\choose k}$$ $${n\choose k+1}=\frac{n-k}{k+1}{n\choose k}$$ If $$\displaystyle{n\choose k}<{n\choose k+1}$$ then $$\implies \frac{n-k}{k+1}>1 \\\implies k<\frac{n-1}2\le\bigg\lfloor\frac n2\bigg\rfloor$$
In the same way we can prove that $$\displaystyle{n\choose k}>{n\choose k+1}$$ only if $$\displaystyle k>\bigg\lceil\frac n2\bigg\rceil$$.
Let me give you an hint by which you can prove it by yourself. Now also let, $$C_{n}=\frac{1}{n+1}\binom{2n}{n}$$ $$\mathbf{CLAIM}:-$$ $$C_{n}=\binom{2n}{n}-\binom{2n}{n+1}$$ $$\mathbf{PROOF}:-$$ $$\binom{2n}{n+1}=\frac{(2n)!}{(n+1)!(n-1)!}=\frac{n}{n+1}\frac{(2n)!}{(n!)^{2}}$$ $$\binom{2n}{n+1}=\frac{n}{n+1}\binom{2n}{n}$$ So $$\binom{2n}{n}-\binom{2n}{n+1}=\frac{1}{n+1}\binom{2n}{n}=C_{n}$$ Hence proved the claim. Now note that $$C_n\ge1$$ so $$\binom{2n}{n}>\binom{2n}{n+1}$$. I think this is enough for you to proceed and prove your required Inequality. The coefficients $$C_{n}$$ that I mentioned are called as Catalan's number. Here's a link if you are interested https://en.m.wikipedia.org/wiki/Catalan_number