1. Find the value of
$$
\binom{6}{1} 2^{1}+\binom{6}{2} 2^{2}+\binom{6}{3} 2^{3}+\binom{6}{4} 2^{4}+\binom{6}{5} 2^{5}+\binom{6}{6} 2^{6} .
$$
Solution:
728
This sum is the binomial expansion of $(1+2)^{6}$, except that it is missing the first term, $\binom{6}{0} 2^{0}=1$. So we get $3^{6}-1=728$.