Question

Find the sum of 0.2 + 0.22 + 0.222 +…….. to n terms.
(a)\[\left( \dfrac{2}{9} \right)-\left( \dfrac{2}{81} \right)\left( 1-{{10}^{-n}} \right)\]
(b)\[n-\left( \dfrac{1}{9} \right)\left( 1-{{10}^{-n}} \right)\]
(c)\[\left( \dfrac{2}{9} \right)\left[ n-\left( \dfrac{1}{9} \right)\left( 1-{{10}^{-n}} \right) \right]\]
(d)\[\left( \dfrac{2}{9} \right)\]

Answer 1

Hint: Take 2 common from series. Multiply and divide by 9 in the series and simplify the series. By using the formula for the sum of n terms in GP, find the sum of n-terms.

Complete step-by-step answer:
Here we need to find the sum of 0.2 + 0.22 + 0.222 +……..+ n terms.
Let us take 2 common from the series.
\[2\left[ 0.1+0.11+0.111+.....+n \right]\]
Now let us multiply and divide by 9 on the side.
\[=\dfrac{2}{9}\left[ 0.9+0.99+0.999+.....+n \right]\]
Now we can modify the series as,
\[\begin{align}
  & 0.9=\dfrac{9}{10}=\dfrac{10-1}{10}=1-\dfrac{1}{10}=1-0.1 \\
 & 0.99=\dfrac{99}{100}=\dfrac{100-1}{100}=1-\dfrac{1}{100}=1-0.01 \\
\end{align}\]
Thus we can write the series as,
\[\dfrac{2}{9}\left[ \left( 1-\dfrac{1}{10} \right)+\left( 1-\dfrac{1}{100} \right)+\left( 1-\dfrac{1}{1000} \right)+.....+n \right]\]
1 is common in all terms, so we can take it out.
\[=\dfrac{2}{9}\left[ \left( 1+1+1+....n \right)-\left( \dfrac{1}{10}+\dfrac{1}{100}+\dfrac{1}{1000}+...+\dfrac{1}{{{10}^{n}}} \right) \right]\]
We know, 1 + 1 + 1 +……. n is n.
\[=\dfrac{2}{9}\left[ n-\left( \dfrac{1}{10}+\dfrac{1}{100}+\dfrac{1}{1000}+....+\dfrac{1}{{{10}^{n}}} \right) \right]-(1)\]
We know that \[\dfrac{1}{10}+\dfrac{1}{100}+\dfrac{1}{1000}+....+\dfrac{1}{{{10}^{n}}}\]is in Geometric Progression.
The GP is a sequence in which each term is derived by multiplying or dividing the preceding term by a fixed number called common ratio, which is denoted by ‘r’.
\[\therefore \]Sum of Geometric Progression is given by the equation,
\[{{S}_{n}}=\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}\]
Considering the GP, \[\dfrac{1}{10}+\dfrac{1}{100}+\dfrac{1}{1000}+....+\dfrac{1}{{{10}^{n}}}\], here first term, \[a=\dfrac{1}{10}\]and \[r=\dfrac{1}{10}\].
\[\therefore \]Sum of n-terms, \[{{S}_{n}}=\dfrac{\dfrac{1}{10}\left[ 1-{{\left( \dfrac{1}{10} \right)}^{n}} \right]}{1-\dfrac{1}{10}}=\dfrac{\dfrac{1}{10}\left[ -{{\left( \dfrac{1}{10} \right)}^{n}} \right]}{10-\dfrac{1}{10}}\]
\[{{S}_{n}}=\dfrac{\left[ 1-{{\left( \dfrac{1}{10} \right)}^{n}} \right]}{9}=\dfrac{1}{9}\left[ 1-{{\left( \dfrac{1}{10} \right)}^{n}} \right]\]
Substitute the value of \[{{S}_{n}}\]in equation (1).
\[\begin{align}
  & =\dfrac{2}{9}\left[ n-\dfrac{1}{9}\left( 1-{{\left( \dfrac{1}{10} \right)}^{n}} \right) \right] \\
 & =\dfrac{2}{9}\left[ n-\dfrac{1}{9}\left( 1-{{10}^{-n}} \right) \right] \\
\end{align}\]
Hence, the sum of series 0.2 + 0.22 + 0.222 +…… + n is equal to \[\dfrac{2}{9}\left[ n-\dfrac{1}{9}\left( 1-{{10}^{-n}} \right) \right]\].
Hence, option (c) is the correct answer.

Note: To solve a question like this we should know the formula connecting to the sum of n –terms in Geometric Progression. Remember to multiply and divide by 9 in the beginning of series as they help us to break down the series.