Answer
Geometric
Sum: $6\left(1-\dfrac{1}{2^n}\right)$
Work Step by Step
We are given the sequence:
$3,\dfrac{3}{2},\dfrac{3}{4},\dfrac{3}{8},\dfrac{3}{16},.....$
We determine the ratio between consecutive terms:
$\dfrac{a_2}{a_1}=\dfrac{\dfrac{3}{2}}{3}=\dfrac{1}{2}$
$\dfrac{a_3}{a_2}=\dfrac{\dfrac{3}{4}}{\dfrac{3}{2}}=\dfrac{1}{2}$
$\dfrac{a_4}{a_3}=\dfrac{\dfrac{3}{8}}{\dfrac{3}{4}}=\dfrac{1}{2}$
$\dfrac{a_5}{a_4}=\dfrac{\dfrac{3}{16}}{\dfrac{3}{8}}=\dfrac{1}{2}$
As the ratio between consecutive terms is constant, the sequence is geometric. Its elements are:
$a_1=3$
$r=\dfrac{1}{2}$
Determine the sum of the first $n$ terms:
$S_n=a_1\cdot\dfrac{1-r^n}{1-r}=3\cdot\dfrac{1-\left(\dfrac{1}{2}\right)^n}{1-\dfrac{1}{2}}=3\left(1-\dfrac{1}{2^n}\right)\cdot 2=6\left(1-\dfrac{1}{2^n}\right)$
