Aloha :)
$$\sum\limits_{k=0}^n\binom{n}{k}\,k\,4^k=\underbrace{\binom{n}{0}\cdot0\cdot4^0}_{=0}+\sum\limits_{k=1}^n\binom{n}{k}\,k\,4^k=\sum\limits_{k=1}^n\frac{n}{\cancel k}\binom{n-1}{k-1}\,\cancel k\,4^k$$$$=n\sum\limits_{k=1}^n\binom{n-1}{k-1} 4^k=n\sum\limits_{k=0}^{n-1}\binom{n-1}{k} 4^{k+1}=4n\sum\limits_{k=0}^{n-1}\binom{n-1}{k} 4^k\cdot1^{(n-1)-k}$$$$=4n(4+1)^{n-1}=4n\,5^{n-1}=\frac{4}{5}\,n\,5^n$$
