Question

The sum of the series $\dfrac{9}{{19}} + \dfrac{{99}}{{{{19}^2}}} + \dfrac{{999}}{{{{19}^3}}} + \dfrac{{9999}}{{{{19}^4}}} + ....\infty$ is
$A)\dfrac{{19}}{{18}}$
$B)\dfrac{{18}}{{19}}$
$C)\dfrac{7}{{18}}$
$D)$ None of these

Answer 1

While talking about the sum of the series, we need to know about the concept of Arithmetic and Geometric progression.

An arithmetic progression can be represented by $a,(a + d),(a + 2d),(a + 3d),...$ where $a$ is the first term and $d$ is a common difference.
A geometric progression can be given by $a,ar,a{r^2},....$ where $a$ is the first term and $r$ is a common ratio.
Hence the given question is in the form of geometric progression.
But here in this question, there is no common difference and thus we cannot apply the AP and GP methods.
We need to solve this in general assumption, we solve step by step using the given information we will conclude the result.

Complete step-by-step solution:
Since from the given that we have, the sum of the given series is $\dfrac{9}{{19}} + \dfrac{{99}}{{{{19}^2}}} + \dfrac{{999}}{{{{19}^3}}} + \dfrac{{9999}}{{{{19}^4}}} + ....\infty$ where it will go to infinity which means undetermined term.
Let us take $S = \dfrac{9}{{19}} + \dfrac{{99}}{{{{19}^2}}} + \dfrac{{999}}{{{{19}^3}}} + \dfrac{{9999}}{{{{19}^4}}} + ....\infty$ then divide $19$ into both sides we get $\dfrac{S}{{19}} = \dfrac{1}{{19}}[\dfrac{9}{{19}} + \dfrac{{99}}{{{{19}^2}}} + \dfrac{{999}}{{{{19}^3}}} + \dfrac{{9999}}{{{{19}^4}}} + ....\infty ]$
Giving the division to each and every term, we get $\dfrac{S}{{19}} = \dfrac{9}{{{{19}^2}}} + \dfrac{{99}}{{{{19}^3}}} + \dfrac{{999}}{{{{19}^4}}} + \dfrac{{9999}}{{{{19}^5}}} + ....\infty$
Now subtract the two equations, which is $S - \dfrac{S}{{19}} = \dfrac{9}{{19}} + \dfrac{{99}}{{{{19}^2}}} + \dfrac{{999}}{{{{19}^3}}} + \dfrac{{9999}}{{{{19}^4}}} + ....\infty - [\dfrac{9}{{{{19}^2}}} + \dfrac{{99}}{{{{19}^3}}} + \dfrac{{999}}{{{{19}^4}}} + \dfrac{{9999}}{{{{19}^5}}} + ....\infty ]$
Combine the terms of the common value, like the same power terms, then we get $S - \dfrac{S}{{19}} = \dfrac{9}{{19}} + (\dfrac{{99}}{{{{19}^2}}} - \dfrac{9}{{{{19}^2}}}) + (\dfrac{{999}}{{{{19}^3}}} - \dfrac{{99}}{{{{19}^3}}}) + (\dfrac{{9999}}{{{{19}^4}}} - \dfrac{{999}}{{{{19}^4}}}) + ....\infty$
Simplifying further in the right and left side, we get $\dfrac{{19S - S}}{{19}} = \dfrac{9}{{19}} + (\dfrac{{90}}{{{{19}^2}}}) + (\dfrac{{900}}{{{{19}^3}}}) + (\dfrac{{9000}}{{{{19}^4}}}) + ....\infty$
Taking the $\dfrac{9}{{19}}$common values out then we get $\dfrac{{18S}}{{19}} = \dfrac{9}{{19}}[1 + (\dfrac{{10}}{{19}}) + (\dfrac{{100}}{{{{19}^2}}}) + (\dfrac{{1000}}{{{{19}^3}}}) + ....\infty ]$ which can be writing as in the form of $\dfrac{{18S}}{{19}} = \dfrac{9}{{19}}[1 + (\dfrac{{10}}{{19}}) + (\dfrac{{100}}{{{{19}^2}}}) + (\dfrac{{1000}}{{{{19}^3}}}) + ....\infty ] \Rightarrow \dfrac{{18S}}{{19}} = \dfrac{9}{{19}}(\dfrac{1}{{1 - \dfrac{{10}}{{19}}}})$
Further solving we get, $\dfrac{{18S}}{{19}} = \dfrac{9}{{19}}(\dfrac{1}{{1 - \dfrac{{10}}{{19}}}}) \Rightarrow \dfrac{9}{{19}} \times \dfrac{{19}}{9} = 1$
Hence, we get $\dfrac{{18S}}{{19}} = 1 \Rightarrow S = \dfrac{{19}}{{18}}$
Therefore, the option $A)\dfrac{{19}}{{18}}$ is correct.

Note: Since we make of the concept of the binomial expansion, which is $1 + \dfrac{1}{2} + {(\dfrac{1}{2})^2} + ....\infty = \dfrac{1}{{1 - \dfrac{1}{2}}}$ and by using this we solved $\dfrac{9}{{19}}[1 + (\dfrac{{10}}{{19}}) + (\dfrac{{100}}{{{{19}^2}}}) + (\dfrac{{1000}}{{{{19}^3}}}) + ....\infty ] = \dfrac{9}{{19}}(\dfrac{1}{{1 - \dfrac{{10}}{{19}}}})$ where this can be rewritten as $\dfrac{9}{{19}}[1 + (\dfrac{{10}}{{19}}) + {(\dfrac{{10}}{{19}})^2} + {(\dfrac{{10}}{{19}})^3} + ....\infty ] = \dfrac{9}{{19}}(\dfrac{1}{{1 - \dfrac{{10}}{{19}}}})$
If there is any common difference in the series, then we will use the AP, GP method to solve.