Answer
$\dfrac{65}{8}$
Work Step by Step
Re-write the given infinite series $\Sigma_{k=0}^{\infty} [\dfrac{1}{2} (0.2)^{k}+\dfrac{3}{2}(0.8)^{k}]$ as $\dfrac{1}{2} \Sigma_{k=0}^{\infty}
(0.2)^{k}+ \dfrac{3}{2} \Sigma_{k=0}^{\infty} (0.8)^{k}$
Here, we can see that the series $\dfrac{1}{2} \Sigma_{k=0}^{\infty}
(0.2)^{k}+ \dfrac{3}{2} \Sigma_{k=0}^{\infty} (0.8)^{k}$
shows a geometric series with common ratio $r=0.2\lt 1$ and $r=0.8 \lt 1$. So, the series is convergent.
whose sum can be computed as: $\Sigma_{n=0}^{\infty} ar^n =a+ar+ar^2+ar^3+.....=\dfrac{a}{1-r}$
Now, $\dfrac{1}{2} \Sigma_{k=0}^{\infty}
(0.2)^{k}+ \dfrac{3}{2} \Sigma_{k=0}^{\infty} (0.8)^{k}
=\dfrac{1}{2} \times \dfrac{1}{1-0.2}+\dfrac{3}{2} \times \dfrac{1}{1-0.8} \\= \dfrac{5}{8}+ \dfrac{15}{2}\\=\dfrac{65}{8}$
