[Ne[S]zA]

$$\sum_{k = 1}^{n} \sum_{i = 1}^{k} \dfrac{1}{3^{i-1}}=\sum_{k = 1}^{n}\dfrac{3}{2}(1-\dfrac{1}{3^k})=\dfrac{3}{2}\sum_{k = 1}^{n}(1-\dfrac{1}{3^k})=\dfrac{3}{2}n-\dfrac{3}{4}(1-\dfrac{1}{3^n})=\dfrac{3^{1-n}+6n-3}{4}$$