Q96 / 170 · Manipal (MET) 2017 · Mathematics
$0.2+0.22+0.222+\ldots$. to $n$ terms is equal to
- A$\left(\frac{2}{9}\right)-\left(\frac{2}{81}\right)\left(1-10^{-n}\right)$
- B$n-\left(\frac{1}{9}\right)\left(1-10^{-n}\right)$
- Correct answerC$\left(\frac{2}{9}\right)\left[n-\left(\frac{1}{9}\right)\left(1-10^{-n}\right)\right]$
- D$\left(\frac{2}{9}\right)$
Solution
$0.2+0.22+0.222+\ldots . n$ terms
$=2(011+011+0.111+\ldots n$ terms $)$
$\begin{aligned} & =2\left(\frac{1}{10}+\frac{11}{100}+\frac{111}{1000}+\ldots n \text { terms }\right) \\ & =\frac{2}{9}\left(\frac{9}{10}+\frac{99}{100}+\frac{999}{1000}+\ldots n \text { terms }\right) \\ & =\frac{2}{9}\left(1-\frac{1}{10}+1-\frac{1}{100}+1-\frac{1}{1000}+\ldots n \text { terms }\right) \\ & =\frac{2}{9}\left[n-\left(\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+\ldots n\right)\right] \\ & =\frac{2}{9}\left[n-\frac{1}{10} \frac{\left\{1-\left(\frac{1}{10}\right)^n\right\}}{\left.\left(1-\frac{1}{10}\right)\right]}\right] \\ & =\frac{2}{9}\left[n-\frac{1}{10} \times \frac{10}{9} \cdot\left(\frac{10^n-1}{10^n}\right)\right] \\ & =\frac{2}{9}\left[n-\frac{1}{9}\left(1-10^{-n}\right)\right]\end{aligned}$
Hence, the correct answer is option: 'C'.