Answer
$\dfrac{11}{2}$
Work Step by Step
The sum of a geometric series can be found as:
$S_n=\dfrac{a_1}{1-r}$
where, $a_1$ denotes the first term and the $r$ is common ratio.
We are given that $\Sigma_{n=0}^{\infty} [(0.6)^n +(0.8)^n]$
This can be further written as: $\Sigma_{n=0}^{\infty} [(0.6)^n +(0.8)^n]=\Sigma_{n=0}^{\infty} (0.6)^n +\Sigma_{n=0}^{\infty} (0.8)^n$
So, we have $a_1=0.6$ and $r_1=0.6$ and $a_2=0.8$ and $r_1=0.8$
Thus, the total sum of two geometric series is:
$S_n=\dfrac{0.6}{1-0.6}+\dfrac{0.8}{1-0.8}=\dfrac{11}{2}$
