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1. Use question $14$ and Corollary $1$ to show that if $n$ is an integer greater than $1,$ then $\binom{n}{\left \lfloor n/2 \right \rfloor}\geq \frac{2^{n}}{2}.$
2. Conclude from part $(A)$ that if $n$ is a positive integer, then $\binom{2n}{n}\geq \frac{4^{n}}{2n}.$

a) By Question 14, we know that  $\binom{n}{\left \lfloor n/2 \right \rfloor}$ is the largest of the $n - 1$ binomial coefficients $\binom{n}{1}$ through $\binom{n}{n-1}$ Therefore it is at least as large as their average, which is $\left ( 2^{n} -2 \right )/\left ( n-1 \right )$ But since $2n\leqslant 2^{_{n}}$ for $n\geqslant 2$ , it follows that  $\left ( 2^{n} -2 \right )/\left ( n-1 \right ) \geqslant 2^{n}/n$, and the proof is complete.
b) This follows from part (a) by replacing $n$ with $2n$ when $n\geq 2$, and it is immediate when $n=1$.