To find the sum of the infinite geometric series [tex]\[-3 - \frac{3}{2} - \frac{3}{4} - \frac{3}{8} - \frac{3}{16} - \ldots,\][/tex] we need to use the formula for the sum of an infinite geometric series. The formula for the sum [tex]\( S \)[/tex] of an infinite geometric series with the first term [tex]\( a \)[/tex] and a common ratio [tex]\( r \)[/tex] is given by:
[tex]\[
S = \frac{a}{1 - r}
\][/tex]
First, let's identify the first term [tex]\( a \)[/tex] and the common ratio [tex]\( r \)[/tex] of the series.
The first term [tex]\( a \)[/tex] is the first number in the series:
[tex]\[
a = -3
\][/tex]
Next, we need to determine the common ratio [tex]\( r \)[/tex]. This can be found by dividing the second term by the first term:
[tex]\[
r = \frac{-\frac{3}{2}}{-3} = \frac{1}{2}
\][/tex]
However, since each term in the series alternates in sign, the correct common ratio is actually:
[tex]\[
r = -\frac{1}{2}
\][/tex]
Now that we have [tex]\( a \)[/tex] and [tex]\( r \)[/tex], we can substitute them into the formula for the sum of an infinite geometric series:
[tex]\[
S = \frac{a}{1 - r} = \frac{-3}{1 - \left(-\frac{1}{2}\right)}
\][/tex]
Simplify the denominator:
[tex]\[
1 - \left(-\frac{1}{2}\right) = 1 + \frac{1}{2} = \frac{2}{2} + \frac{1}{2} = \frac{3}{2}
\][/tex]
Now substitute this back into the formula:
[tex]\[
S = \frac{-3}{\frac{3}{2}} = -3 \times \frac{2}{3} = -2
\][/tex]
Thus, the sum of the infinite geometric series is:
[tex]\[
\boxed{-2}
\][/tex]
