Question

The sum of the series$\frac{9}{19}+\frac{99}{{19}^2}+\frac{999}{{19}^3}+.............\infty$is

• A. $\frac{19}{18}$
• B. $\frac{18}{19}$
• C. $\frac{7}{18}$
• D. None of these

Answer 1

The correct option is A

$\frac{19}{18}$

Explanation for the correct option

Given: $\frac{9}{19}+\frac{99}{{19}^2}+\frac{999}{{19}^3}+.............\infty$

Let $S=\frac{9}{19}+\frac{99}{{19}^2}+\frac{999}{{19}^3}+.............\infty$ -----------(1)

Divide both sides by $19$

$\therefore \frac{S}{19}=\frac{9}{{19}^2}+\frac{99}{{19}^3}+\frac{999}{{19}^4}+.............\infty$ ------------(2)

Subtracting (2) from (1)

$$\begin{gathered} \therefore S-\frac{S}{19}=\frac{9}{19}+\frac{99}{{19}^2}+\frac{999}{{19}^3}+............-\left(\frac{9}{{19}^2}+\frac{99}{{19}^3}+\frac{999}{{19}^4}+...........\right) \\ \Rightarrow \frac{18}{19}S=\frac{9}{19}+\left(\frac{99}{{19}^2}-\frac{9}{{19}^2}\right)+\left(\frac{999}{{19}^3}-\frac{99}{{19}^3}\right)+........... \\ \Rightarrow \frac{18}{19}S=\frac{9}{19}+\frac{90}{{19}^2}+\frac{900}{{19}^3}+........... \\ \Rightarrow \frac{18}{19}S=\frac{9}{19}\left(1+\frac{10}{19}+\frac{100}{{19}^2}+.........\right) \end{gathered}$$

We know that ${(1-x)}^{-1}=1+x+x^2+x^3+.......$

$\therefore$${\left(1-\frac{10}{19}\right)}^{-1}=1+\frac{10}{19}+{\left(\frac{10}{19}\right)}^2+{\left(\frac{10}{19}\right)}^3+.......$

$$\begin{gathered} \Rightarrow \frac{9}{19}\left(1+\frac{10}{19}+\frac{100}{{19}^2}+.........\right)=\frac{9}{19}{\left(1-\frac{10}{19}\right)}^{-1} \\ \Rightarrow \frac{18}{19}S=\frac{9}{19}{\left(\frac{9}{19}\right)}^{-1} \\ \Rightarrow \frac{18}{19}S=\frac{9}{19}\times \frac{19}{9} \\ \Rightarrow S=\frac{19}{18} \end{gathered}$$

Hence, $\mathrm{Option}\left(\mathrm{A}\right)$ is correct.