Just some random sum

Let $\displaystyle S = \frac {1\times 2}3 + \frac{2 \times 3}{3^2} + \frac {3 \times 4}{3^3} + \frac {4\times 5}{3^4} + ...$ , then let us consider the following.

$\begin{aligned} \frac 1{1-x} & = 1 + x + x^2 + x^3 + ... & \small \color{#3D99F6}{\text{Multiply both sides by }x^2} \\ \frac {x^2}{1-x} & = x^2 + x^3 + x^4 + x^5 + ... & \small \color{#3D99F6}{\text{Differentiate w.r.t. }x} \\ \frac {2x-x^2}{(1-x)^2} & = 2x + 3x^2 + 4x^3 + 5x^4 + ... & \small \color{#3D99F6}{\text{Differentiate w.r.t. }x \text{ again}} \\ \frac 2{(1-x)^3} & = 1\times 2 + 2 \times 3x + 3 \times 4x^2 + 4\times 5x^3 + ... & \small \color{#3D99F6}{\text{Multiply by }x} \\ \frac {2x}{(1-x)^3} & = 1\times 2x + 2 \times 3x^2 + 3 \times 4x^3 + 4\times 5x^4 + ... & \small \color{#3D99F6}{\text{Put }x = \frac 13} \\ \frac {\frac 23}{\left(1-\frac 13\right)^3} & = \frac {1\times 2}3 + \frac{2 \times 3}{3^2} + \frac {3 \times 4}{3^3} + \frac {4\times 5}{3^4} + ... \color{#3D99F6}{= S} \end{aligned}$

$\implies S = \dfrac {\frac 23}{\left(1-\frac 13\right)^3} = \dfrac 23 \times \left(\dfrac {3}{2}\right)^3 = \boxed{\dfrac 94}$