How to prove $\binom{n}{n/2} = \Omega (\frac{2^{n}}{n})$?
Hint: What is the sum of $\binom{n}{k}$ for all $k$? And which is the greatest of these?
An alternative is to use Stirling's formula to get $$\binom{n}{n/2} = \Theta\left(\frac{2^n}{\sqrt{n}}\right).$$
You can also get the same result from the Central Limit Theorem, since $$\frac{\binom{n}{n/2}}{2^n} = \mathrm{Pr}[\mathrm{Bin(n,\frac{1}{2})}=\frac{n}{2}].$$