[tex]\qquad \qquad \textit{sum of a finite geometric sequence} \\\\ \displaystyle S_n=\sum\limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad \begin{cases} n=\textit{last term's}\\ \qquad position\\ a_1=\textit{first term}\\ r=\textit{common ratio} \end{cases}[/tex]
let's get the summation of the first 14 terms and then the summation of the first 2 terms, and subtract the later from the former, leaving us only with the summation from 3 through 14.
[tex]A(n)=\stackrel{a_1}{4}(\underset{r}{3})^{n-1}\implies \displaystyle\sum_{i=1}^{14}~~4(3)^{n-1}~~ -~~\sum_{i=1}^{2}~~4(3)^{n-1} \\\\\\ 4\left( \cfrac{1-3^{14}}{1-3} \right)~~ - ~~4\left( \cfrac{1-3^{2}}{1-3} \right)\implies 4(2391484)~~ - ~~4(4) \\\\\\ 9565936~~ -~~16\implies \boxed{9565920}[/tex]
