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

Answer 1

The correct option is C $\left(\frac{2}{9}\right)[n-\left(\frac{1}{9}\right)(1-{10}^{-n}\left)\right]$

General term is, $t_r=0.222222222(r$ times )

$=\frac{2}{10}+\frac{2}{{10}^2}+\frac{2}{{10}^3}+....+\frac{2}{{10}^r}=\frac{2}{10}\cdot \frac{1-{10}^{-r}}{1-\frac{1}{10}}=\frac{2}{9}(1-{10}^{-r})$

Hence required sum $=\sum _{r=1}^n{t}_r=\frac{2n}{9}-\frac{2}{9}\sum _{r=1}^n{10}^{-r}$

$=\frac{2n}{9}-\frac{2}{9}\cdot \frac{{10}^{-1}(1-{10}^{-n})}{1-1/10}$

$=\frac{2}{9}[n-\frac{1}{9}(1-{10}^{-n}\left)\right]$