Answer
$$\sum\limits_{n = 1}^\infty {\left( {\frac{3}{2}} \right){{\left( {\frac{3}{4}} \right)}^{n - 1}}} $$
Work Step by Step
$$\eqalign{
& {\text{Let }}r = \frac{3}{4}{\text{ and }}{S_\infty } = 6 \cr
& {\text{The Sum of the Terms of an Infinite Geometric Sequence is}} \cr
& {S_\infty } = \frac{{{a_1}}}{{1 - r}} \cr
& 6 = \frac{{{a_1}}}{{1 - 3/4}} \cr
& 6 = \frac{{{a_1}}}{{1/4}} \cr
& {a_1} = \frac{3}{2} \cr
& {\text{The general term of the geometric series is }} \cr
& {a_n} = {a_1}{r^{n - 1}} \cr
& {a_n} = \left( {\frac{3}{2}} \right){\left( {\frac{3}{4}} \right)^{n - 1}} \cr
& {\text{The infinite geometric series is}} \cr
& \sum\limits_{n = 1}^\infty {\left( {\frac{3}{2}} \right){{\left( {\frac{3}{4}} \right)}^{n - 1}}} \cr} $$
