Answer
$\dfrac{27}{5}$
Work Step by Step
Re-write the given infinite series $\Sigma_{k=1}^{\infty} [2 (\dfrac{3}{5})^{k}+3(\dfrac{4}{9})^{k}]$ as $2 \Sigma_{k=1}^{\infty} (\dfrac{3}{5})^{k}+3 \Sigma_{k=1}^{\infty} (\dfrac{4}{9})^{k}$
Here, we can see that the series $2 \Sigma_{k=1}^{\infty} (\dfrac{3}{5})^{k}+3 \Sigma_{k=1}^{\infty} (\dfrac{4}{9})^{k}$ shows a geometric series with common ratio $r=\dfrac{3}{5}\lt 1$ and $r=\dfrac{4}{9} \lt 1$. 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, $2 \Sigma_{k=1}^{\infty} (\dfrac{3}{5})^{k}+3 \Sigma_{k=1}^{\infty} (\dfrac{4}{9})^{k}=2 \times \dfrac{3/5}{1-\dfrac{3}{5}}+3 \times \dfrac{4/9}{1-\dfrac{4}{9}} \\=2 \times \dfrac{3}{2}+3 \times \dfrac{4}{5}\\=\dfrac{27}{5}$
