Answer
$-2$
Work Step by Step
Re-write the given infinite series $\Sigma_{k=0}^{\infty} [3 (\dfrac{2}{5})^{k}-2(\dfrac{5}{7})^{k}]$ as $3 \Sigma_{k=0}^{\infty} (\dfrac{2}{5})^{k}-2 \Sigma_{k=0}^{\infty} (\dfrac{5}{7})^{k}$
Here, we can see that the series $3 \Sigma_{k=0}^{\infty} (\dfrac{2}{5})^{k}-2 \Sigma_{k=0}^{\infty} (\dfrac{5}{7})^{k}$ shows a geometric series with common ratio $r=\dfrac{2}{5}\lt 1$ and $r=\dfrac{5}{7} \lt 1$. So, the series is convergent.
whose sum can be computed as: $\Sigma_{n=1}^{\infty} a^n =a+a^2+a^3+.....=\dfrac{a}{1-a}$
Now, $3 \Sigma_{k=0}^{\infty} (\dfrac{2}{5})^{k}-2 \Sigma_{k=0}^{\infty} (\dfrac{5}{7})^{k}=3 \times \dfrac{1}{1-\dfrac{2}{5}}- 2 \times \dfrac{1}{1-\dfrac{5}{7}} \\=3 \times \dfrac{5}{3} -2 \times \dfrac{7}{2}\\=5-7\\=-2$
