The sum of the infinite series $$\cfrac{1}{5}\left(\cfrac{1}{5} - \cfrac{1}{7}\right) + \left(\cfrac{1}{5}\right)^2 \left(\left(\cfrac{1}{5}\right)^2 - \left(\cfrac{1}{7}\right)^2\right) + \left(\cfrac{1}{5}\right)^3 \left(\left(\cfrac{1}{5}\right)^3 - \left(\cfrac{1}{7}\right)^3\right) + ......$$ is equal to

Opening the brackets, we get the series as: $$\left(\frac{1}{5}\right)^2-\left(\frac{1}{5}\times\ \frac{1}{7}\right)+\left(\frac{1}{5}\right)^4-\left(\frac{1}{5}\times\ \frac{1}{7}\right)^2+\left(\frac{1}{5}\right)^6-\left(\frac{1}{5}\times\ \frac{1}{7}\right)^6+...$$

These are two infinite GPs when rearranged:
$$\left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^4+\left(\frac{1}{5}\right)^6+...-\left(\frac{1}{5}\times\ \frac{1}{7}\right)-\left(\frac{1}{5}\times\ \frac{1}{7}\right)^6-\left(\frac{1}{5}\times\ \frac{1}{7}\right)^2-...$$

The sum of the first series would be $$\frac{\frac{1}{25}}{1-\frac{1}{25}}=\frac{1}{24}$$
The sum of the second series would be $$\frac{\frac{1}{35}}{1-\frac{1}{35}}=\frac{1}{34}$$

The answer to the given series would then be $$\frac{1}{24}-\frac{1}{34}=\frac{10}{816}=\frac{5}{408}$$

Therefore, Option B is the correct answer.