Để $2^kC_{30}^k$ là $\max$ thì
$\begin{cases} 2^kC_{30}^k \ge 2^{k+1} C_{30}^{k+1} \ (1)\\ 2^kC_{30}^k \ge 2^{k-1} C_{30}^{k-1} \ (2)\end{cases}$
$(1) \Leftrightarrow 2^k \dfrac{30!}{k! (30-k)!} \ge 2^{k+1} \dfrac{30!}{(k+1)! (29-k)!} $
$\Leftrightarrow \dfrac{1}{30-k} \ge 2 . \dfrac{1}{k+1} \ (*)$
$(2) \Leftrightarrow 2^k \dfrac{30!}{k! (30-k)!} \ge 2^{k-1} \dfrac{30!}{(k-1)! (31-k)!} $
$\Leftrightarrow \dfrac{1}{k} \ge \dfrac{1}{2 (31-k)} \ (**)$
Giải $(*);\ (**) \Rightarrow \dfrac{59}{3}\ \le k \ \le \dfrac{62}{3};\ k\in Z \Rightarrow k = 20$