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

S =$$\frac{3}{7}+\frac{4}{7^2}+\frac{5}{7^3}+\frac{3}{7^4}+\frac{4}{7^5}+\frac{5}{7^6}+......$$
Now here S is a summation of 3 distinct infinite geometric series
And we know that sum of an infinite geometric progression = (a/1-r) where a is the first term and r is the common ratio .
Therefore we can say S will be : $$\frac{3}{\frac{7}{1-\frac{1}{7^3}}}+\ \frac{4}{\frac{7^2}{1-\frac{1}{7^3}}}+\frac{5}{\frac{7^3}{1-\frac{1}{7^3}}}$$
solving we get S = $$\frac{180}{7^3-1}=\frac{180}{342}$$