21. Find the sum $\sum_{k=1}^{19} k\binom{19}{k}$.
(ans. $19 \cdot 2^{18}$.
$\binom{n}{k}=\frac{n}{k}\binom{n-1}{k-1} \Rightarrow k\binom{n}{k}=n\binom{n-1}{k-1}$. Then, $n=19 \Rightarrow$ $\left.\sum_{k=1}^{19} k\binom{19}{k}=\sum_{k=0}^{18} 19\binom{18}{k-1}=19 \cdot 2^{18}.\right)$