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 $n=1$.