The sum of the infinite series $1+\frac{5}{6}+\frac{12}{6^2}+\frac{22}{6^3}+\frac{35}{6^4}+\frac{51}{6^5}+\frac{70}{6^6}+\dots$ is equal to:

Given,

$S=1+\frac{5}{6}+\frac{12}{6^2}+\frac{22}{6^3}+\frac{35}{6^4}+\dots .$ (i)

Multiplying with $\frac{1}{6}$ both side

$\frac{S}{6}=\frac{1}{6}+\frac{5}{6^2}+\frac{12}{6^3}+\frac{22}{6^4}+\dots$ (ii)

On subtraction equation (i) - equation (ii)

We get, $\frac{5}{6}S=1+\frac{4}{6}+\frac{7}{6^2}+\frac{10}{6^3}+\frac{13}{6^4}+\dots$(iii)

Again multiplying $\frac{1}{6}$ both side,

$\frac{5}{36}S=1+\frac{4}{6^2}+\frac{7}{6^3}+\frac{10}{6^4}+\frac{13}{6^5}+...$(iv)

On subtraction equation (iii) - equation (iv)

$\frac{25}{36}S=1+\frac{3}{6}+\frac{3}{6^2}+\frac{3}{6^3}+\dots =\frac{8}{5}$

$\mathrm{S}=\frac{288}{125}$