<s>[INST] Proving that for $n \equiv 0 \pmod{2}$, we get ${n \choose 0} < {n \choose 1} <\ldots< {n \choose n/2-1}<{n\choose n/2}$ etc. How can one prove that for $n \equiv 0$ mod $2$  we have
$${n \choose 0} < {n \choose 1} <\ldots< {n \choose n/2-1}<{n \choose n/2}>{n\choose n/2+1}>\ldots>{n\choose n-1}>{n\choose n}\,?$$
Can I say that for fixed $n$, the binomial coefficients $n \choose k$ increase with $k$ for $k < n/2$? If n is even (like in our case), then the central binomial coefficient $n \choose n/2$ is the largest one. 
So $n \choose k+1$ is greater than, equal to, or less than $n \choose k$ according as $n-k$ is greater than, equal to, or less than $k+1$, that is according as $k$ is less than, equal to, or greater than $(n−1)/2$.
Is that correct?
 [/INST] Let's try using induction on $m$ where $n=2m$. For $m=0$, there is nothing to be proven, for $m=1$, then ${2\choose 0} = {2\choose 2} = 1 < 2 = {2\choose 1}$.
To get the induction step, observe that ${n+2\choose k} = {n+1\choose k-1} + {n+1 \choose k} = {n\choose k-2} + 2 {n\choose k-1} + {n\choose k}$. Suppose that $k<m+1 = \frac{n+2}{2}$, then
\begin{align*}
{2(m+1)\choose k} - {2(m+1)\choose k+1} &= {2m \choose k-2}+{2m\choose k-1}-{2m \choose k}-{2m\choose k+1}\\
&= \left[ {2m \choose k-2} - {2m \choose k} \right] + \left[ {2m\choose k-1} -{2m\choose k+1}\right]
\end{align*}
The first term is directly strictly negative by induction, the second term is too if $k<m$, if $k=m$ then ${2m\choose m-1}={2m\choose m+1}$ and the second term is $0$. This proves that ${2n+2\choose k}<{2n+2\choose k+1}$. The same can be done when $k>m+1$ to obtain ${2n+2\choose k}<{2n+2\choose k-1}$.
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