Answer:
[tex]S=-6[/tex]
Step-by-step explanation:
Sum Of A Geometric Series
Given a geometric series
[tex]a_1,\ a_1.r,\ a_1.r^2,\ a_1.r^3,...[/tex]
The sum of the infinite terms is given by
[tex]\displaystyle S_n=\frac{a_1}{1-r}[/tex]
The sum converges only if
[tex]|r|<1[/tex]
We are given the series as a sum:
[tex]S=\displaystyle -3-\frac{3}{2}-\frac{3}{4}-\frac{3}{8}-\frac{3}{16}...[/tex]
Factoring by -3:
[tex]\displaystyle -3\left (1+ \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}...\right )[/tex]
The expression in parentheses is a geometric series with
[tex]a_1=1,\ r=1/2[/tex]
The sum can be computed by using the formula
[tex]\displaystyle S_n=\frac{1}{1-1/2}[/tex]
[tex]S_n=2[/tex]
So our expression becomes
[tex]S=(-3)(2)[/tex]
[tex]\boxed{S=-6}[/tex]
