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,

& 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{aligned}$$ Thus we can write the series as, $$\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{aligned} & =\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{aligned}$$ Hence, the sum of series 0.2 + 0.22 + 0.222 +…… + n is equal to $$\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.