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+....+\text{ }n\text{ }terms $ $ =2(0.1+0.11+0.111+...+n\text{ }terms) $ $ =2\left( \frac{1}{10}+\frac{11}{100}+\frac{111}{1000}+......+n\text{ }terms \right) $ $ =\frac{2}{9}\left( \frac{9}{10}+\frac{99}{100}+\frac{999}{1000}+....+n\text{ }terms \right) $ $ =\frac{2}{9}\left( 1-\frac{1}{10}+1-\frac{1}{100}+1-\frac{1}{1000}+.... \right. $ $ +n\text{ }terms) $ $ =\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] $