Answer
$\dfrac{17}{10}$
Work Step by Step
Re-write the given infinite series $\Sigma_{k=1}^{\infty} [(\dfrac{1}{6} )^{k}+(\dfrac{1}{3})^{k-1}]$ as $\Sigma_{k=1}^{\infty} (\dfrac{1}{6} )^{k}+3 \Sigma_{k=1}^{\infty} (\dfrac{1}{3})^{k}$
Here, we can see that the series $\Sigma_{k=1}^{\infty} (\dfrac{1}{6} )^{k}+3 \Sigma_{k=1}^{\infty} (\dfrac{1}{3})^{k}$ shows a geometric series .
whose sum can be computed as: $\Sigma_{n=0}^{\infty} ar^n =a+ar+ar^2+ar^3+.....=\dfrac{a}{1-r}$
Now, $\Sigma_{k=1}^{\infty} (\dfrac{1}{6} )^{k}+3 \Sigma_{k=1}^{\infty} (\dfrac{1}{3})^{k}
=\dfrac{1/6}{1-1/6}+3 \times \dfrac{1/3}{1-1/3} \\= \dfrac{1}{5}+ \dfrac{3}{2}\\=\dfrac{17}{10}$
