Question

$0.2+0.22+0.222+...$ to n terms is equal to:

• A. $\left(\frac{2}{9}\right)-\left(\frac{2}{81}\right)(1-{10}^{-n})$
• B. $n-\left(\frac{1}{9}\right)(1-{10}^{-n})$
• C. $\left(\frac{2}{9}\right)\left[n-\left(\frac{1}{9}\right)(1-{10}^{-n})\right]$
• D. $\left(\frac{2}{9}\right)$
• E. $\left(\frac{2}{9}\right)\left[n-\left(\frac{1}{9}\right)(1-{10}^n)\right]$

Answer 1

Answer: (C)

$0.2+0.22+0.222+....+nterms$ $=2(0.1+0.11+0.111+...+nterms)$ $=2\left(\frac{1}{10}+\frac{11}{100}+\frac{111}{1000}+......+nterms\right)$ $=\frac{2}{9}\left(\frac{9}{10}+\frac{99}{100}+\frac{999}{1000}+....+nterms\right)$ $=\frac{2}{9}(1-\frac{1}{10}+1-\frac{1}{100}+1-\frac{1}{1000}+....$ $+nterms)$ $=\frac{2}{9}\left[n-\left(\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+....n\right)\right]$ $=\frac{2}{9}\left[n-\frac{1}{10}\frac{\left[1-{\left(\frac{1}{10}\right)}^n\right]}{\left[1-\frac{1}{10}\right]}\right]$ $=\frac{2}{9}\left[n-\frac{1}{10}\times \frac{10}{9}.\left[\frac{{10}^n-1}{{10}^n}\right]\right]$ $=\frac{2}{9}\left[n-\frac{1}{9}(1-{10}^{-n})\right]$