Approach 1: Manipulation
Let sum = S.
$$S=\frac{1}{3^1}+\frac{1}{3^2}+\frac{2}{3^2}+\frac{1}{3^3}+\frac{3}{3^3}+...=\left(\frac{1}{3^1}+\frac{1}{3^2}+...\right)+\frac{S}{3}=\frac{1}{2}+\frac{S}{3}$$
Hence $S=\frac{S}{3}+\frac{1}{2}\implies \frac{2S}{3}=\frac{1}{2}\implies \boxed{S=\frac{3}{4}}$.
Approach 2: Calculus
Take the series $\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n$. Differentiating yields $\frac{1}{(1-x)^2}=\sum_{n=0}^{\infty} nx^{n-1}\implies \frac{x}{(1-x)^2}=\sum_{n=0}^{\infty} nx^n$. Putting $x=\frac{1}{3}$, the answer is $\frac{\frac{1}{3}}{\frac{2}{3}\frac{2}{3}}=\frac{3}{4}$.
