Question

The sum of $0.2+0.22+0.222+.............$ to $n$ terms is equal to

Answer 1

Step 1: Simplify the given expression

Given, $0.2+0.22+0.222+...$

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

Step 2: Simplify the geometric progression

$\frac{1}{10},\frac{1}{100},\frac{1}{1000},...$ is a geometric progression with first term $a=\frac{1}{10}$ and coomon ratio $r=\frac{1}{10}$.

Sum of geometric progression$=a\left[\frac{1-r^n}{1-r}\right]$ , for $r<1$

Sum of $\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+.......$ $=\frac{1}{10}\left[\frac{1-{\left({\displaystyle \frac{1}{10}}\right)}^n}{{\displaystyle 1-\frac{1}{10}}}\right]$

$$\begin{gathered} =\frac{1}{10}\times \left[\frac{1-{\left(\frac{1}{10}\right)}^n}{{\displaystyle \frac{9}{10}}}\right] \\ =\frac{1}{10}\times \frac{10}{9}\left[1-{\left(\frac{1}{10}\right)}^n\right] \\ =\frac{1}{9}\left[1-{\left(\frac{1}{10}\right)}^n\right] \end{gathered}$$

$\Rightarrow \frac{2}{9}\left[n-\left(\frac{1}{10}+\frac{1}{100}+\frac{1}{1000}+.......\right)\right]=\frac{2}{9}\left[n-\frac{1}{9}{\left(1-\frac{1}{10}\right)}^n\right]$

Hence, $0.2+0.22+0.222+...=\frac{2}{9}\left[n-\frac{1}{9}{\left(1-\frac{1}{10}\right)}^n\right]$.