To find the sum of the infinite geometric series [tex]\(\sum_{n=1}^{\infty} (-144) \left(\frac{1}{2}\right)^{n-1}\)[/tex], let's follow the steps below:
1. Identify the first term [tex]\(a\)[/tex] and the common ratio [tex]\(r\)[/tex]:
- The first term [tex]\(a\)[/tex] of the series is [tex]\(-144\)[/tex]. This is because when [tex]\(n=1\)[/tex], the term is [tex]\((-144) \left(\frac{1}{2}\right)^{1-1} = -144 \cdot 1 = -144\)[/tex].
- The common ratio [tex]\(r\)[/tex] is [tex]\(\frac{1}{2}\)[/tex]. Each subsequent term in the series is multiplied by [tex]\(\frac{1}{2}\)[/tex] compared to the previous term.
2. Recall the formula for the sum [tex]\(S\)[/tex] of an infinite geometric series:
The formula for the sum of an infinite geometric series [tex]\(\sum_{n=0}^{\infty}ar^n\)[/tex], where [tex]\(|r| < 1\)[/tex], is given by:
[tex]\[
S = \frac{a}{1 - r}
\][/tex]
3. Substitute the values of [tex]\(a\)[/tex] and [tex]\(r\)[/tex] into the formula:
[tex]\[
S = \frac{-144}{1 - \frac{1}{2}}
\][/tex]
4. Simplify the expression:
[tex]\[
S = \frac{-144}{1 - \frac{1}{2}} = \frac{-144}{\frac{1}{2}} = -144 \cdot \frac{2}{1} = -144 \cdot 2 = -288
\][/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].
