Find the sum of following series:
$$\dfrac{3}{7}+\dfrac{4}{7^2}+\dfrac{5}{7^3}+\dfrac{3}{7^4}+\dfrac{4}{7^5}+\dfrac{5}{7^6}+\dfrac{3}{7^7}+\dfrac{4}{7^8}+\dfrac{5}{7^9}+...$$

S =$$\frac{3}{7}+\frac{4}{7^2}+\frac{5}{7^3}+\frac{3}{7^4}+\frac{4}{7^5}+\frac{5}{7^6}+......$$ 

S is a summation of 3 distinct infinite geometric series as we can see that the terms {1,4,7...}, {2,5,8...}, {3,6,9...} are forming 3 separate infinite GPs.
We know that the sum of an infinite geometric progression = $$\dfrac{a}{1\ -\ r}$$ where a is the first term and r is the common ratio.

Therefore we can say S will be : $$\dfrac{\frac{3}{7}}{1-\frac{1}{7^3}}+\ \dfrac{\frac{4}{7^2}}{1-\frac{1}{7^3}}+\dfrac{\frac{5}{7^3}}{1-\frac{1}{7^3}}$$

solving we get S = $$\dfrac{180}{7^3-1}=\dfrac{180}{342}$$