Answer
$$\sum\limits_{i = 1}^6 {4{{\left( 3 \right)}^{i - 1}}} $$
Work Step by Step
$$\eqalign{
& 4 + 12 + 36 + \cdots 972 \cr
& {\text{Calculating }}d{\text{ or }}r \cr
& 12 - 4 = 8,\,\,\,\,\,\,\,\,\,36 - 12 = 24 \cr
& \frac{{12}}{4} = 3,\,\,\,\,\frac{{36}}{{12}} = 3 \cr
& {\text{This is a geometric series with }}{a_1} = 4{\text{ and }}r = 3 \cr
& {\text{The general term is }} \cr
& {a_n} = {a_1}{r^{n - 1}} \cr
& {a_n} = 4{\left( 3 \right)^{n - 1}} \cr
& {\text{The last term is }}86,{\text{ then}} \cr
& 972 = 4{\left( 3 \right)^{n - 1}} \cr
& {3^{n - 1}} = 243 \cr
& n = 6 \cr
& {\text{The sum of notation can be written as}} \cr
& \sum\limits_{i = 1}^6 {4{{\left( 3 \right)}^{i - 1}}} \cr} $$
