To find the sum of the infinite geometric series [tex]\(\sum_{n=1}^{\infty}(-144)\left(\frac{1}{2}\right)^{n-1}\)[/tex], we need to identify the first term and the common ratio of the series.
1. Identify the first term [tex]\(a\)[/tex]:
The given series can be written in a general form for a geometric series:
[tex]\[
\sum_{n=1}^{\infty} a \cdot r^{n-1}
\][/tex]
Here, the first term [tex]\(a\)[/tex] is [tex]\(-144\)[/tex].
2. Identify the common ratio [tex]\(r\)[/tex]:
The common ratio [tex]\(r\)[/tex] is the factor by which each term is multiplied to get the next term. In this series, [tex]\(r = \frac{1}{2}\)[/tex].
3. Formula for the sum of an infinite geometric series:
The sum [tex]\(S\)[/tex] of an infinite geometric series [tex]\(\sum_{n=1}^{\infty} ar^{n-1}\)[/tex] is given by the formula:
[tex]\[
S = \frac{a}{1 - r}
\][/tex]
This formula applies when the absolute value of the common ratio is less than 1 ([tex]\(|r| < 1\)[/tex]).
4. Substitute the values into the formula:
[tex]\[
a = -144, \quad r = \frac{1}{2}
\][/tex]
Plug these into the formula:
[tex]\[
S = \frac{-144}{1 - \frac{1}{2}}
\][/tex]
5. Simplify the expression:
[tex]\[
S = \frac{-144}{1 - \frac{1}{2}} = \frac{-144}{\frac{1}{2}}
\][/tex]
Dividing by [tex]\(\frac{1}{2}\)[/tex] is equivalent to multiplying by 2:
[tex]\[
S = -144 \times 2 = -288
\][/tex]
Therefore, the sum of the infinite geometric series [tex]\(\sum_{n=1}^{\infty}(-144)\left(\frac{1}{2}\right)^{n-1}\)[/tex] is [tex]\(\boxed{-288}\)[/tex].
