To evaluate the infinite series [tex]\(\sum_{k=0}^{\infty} 6\left(-\frac{2}{3}\right)^k\)[/tex], we recognize that this is an infinite geometric series.
The general form of an infinite geometric series is given by:
[tex]\[ S = a + ar + ar^2 + ar^3 + \cdots, \][/tex]
where [tex]\(a\)[/tex] is the first term and [tex]\(r\)[/tex] is the common ratio.
For our series:
1. The first term [tex]\(a\)[/tex] is [tex]\(6\)[/tex].
2. The common ratio [tex]\(r\)[/tex] is [tex]\(-\frac{2}{3}\)[/tex].
The formula for the sum of an infinite geometric series [tex]\(S\)[/tex] is:
[tex]\[ S = \frac{a}{1 - r}, \][/tex]
provided that [tex]\(|r| < 1\)[/tex]. Here, the common ratio [tex]\(r = -\frac{2}{3}\)[/tex] has an absolute value [tex]\(|-\frac{2}{3}| = \frac{2}{3}\)[/tex], which is less than 1. Hence, we can use the formula.
Let's plug in the values we have:
- [tex]\(a = 6\)[/tex],
- [tex]\(r = -\frac{2}{3}\)[/tex].
So, the sum [tex]\(S\)[/tex] is:
[tex]\[
S = \frac{6}{1 - \left(-\frac{2}{3}\right)}.
\][/tex]
Simplifying inside the denominator:
[tex]\[
1 - \left(-\frac{2}{3}\right) = 1 + \frac{2}{3} = \frac{3}{3} + \frac{2}{3} = \frac{5}{3}.
\][/tex]
Putting this back into our formula:
[tex]\[
S = \frac{6}{\frac{5}{3}}.
\][/tex]
Dividing by a fraction is the same as multiplying by its reciprocal:
[tex]\[
S = 6 \times \frac{3}{5} = \frac{18}{5}.
\][/tex]
To convert [tex]\(\frac{18}{5}\)[/tex] to a decimal:
[tex]\[
\frac{18}{5} = 3.6.
\][/tex]
Therefore, the sum of the infinite series [tex]\(\sum_{k=0}^{\infty} 6\left(-\frac{2}{3}\right)^k\)[/tex] is:
[tex]\[
\boxed{3.6}.
\][/tex]
