(i) (0.2+0.22+0.222+\cdots) to (n) terms

[ S_n = \dfrac{2}{10}+\dfrac{22}{100}+\dfrac{222}{1000}+\cdots+\underbrace{0.222\ldots2}_{n\text{ digits}} ] [ = \dfrac{2}{9}\left[\left(1-\dfrac{1}{10}\right)+\left(1-\dfrac{1}{10^2}\right)+\cdots+\left(1-\dfrac{1}{10^n}\right)\right] ] [ = \dfrac{2}{9}\left[n - \dfrac{1}{10}\cdot\dfrac{1-\left(\frac{1}{10}\right)^n}{1-\frac{1}{10}}\right] = \dfrac{2}{9}\left[n - \dfrac{1}{9}\left(1-\dfrac{1}{10^n}\right)\right] ]

Answer: (\dfrac{2}{9}\left[n-\dfrac{1}{9}\left(1-10^{-n}\right)\right])

(ii) (3+33+333+\cdots) to (n) terms

[ S_n=\dfrac{3}{9}\left[9+99+999+\cdots\right]=\dfrac{1}{3}\sum_{k=1}^{n}(10^k-1) =\dfrac{1}{3}\left[\dfrac{10(10^n-1)}{9}-n\right] ]

Answer: (\dfrac{1}{3}\left[\dfrac{10(10^n-1)}{9}-n\right]=\dfrac{10^{n+1}-10-9n}{27})