The sum of the infinite series 1 + (5)/(6) + 12 6 ^ 2 + 22 6 ^ 3 +...

Given,

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

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

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

On subtraction equation (i) - equation (ii)

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

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

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

On subtraction equation (iii) - equation (iv)

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

$S = \dfrac{288}{125}$