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

$\underline{\begin{array}{r}\mathrm{S}=1+\frac{5}{6}+\frac{12}{6^2}+\frac{22}{6^3}+\frac{35}{6^4}+\dots .\\ \frac{S}{6}=\frac{1}{6}+\frac{5}{6^2}+\frac{12}{6^3}+\frac{22}{6^4}+\dots .\end{array}}$
on subtraction
$\underline{\begin{array}{r}\frac{5}{6}\mathrm{S}=1+\frac{4}{6}+\frac{7}{6^2}+\frac{10}{6^3}+\frac{13}{6^4}+\dots \\ \frac{5}{36}\mathrm{S}=1+\frac{4}{6^2}+\frac{7}{6^3}+\frac{10}{6^4}+\frac{13}{6^5}+\dots \end{array}}$
on subtraction
$\begin{array}{l}\frac{25}{36}\mathrm{S}=1+\frac{3}{6}+\frac{3}{6^2}+\frac{3}{6^3}+\dots =\frac{8}{5}\\ \mathrm{S}=\frac{288}{125}\end{array}$