Question

The sum of infinity of the series $1 + \dfrac{4}{5} + \dfrac{7}{{{5^2}}} + \dfrac{{10}}{{{5^3}}} + .....$ is
A) $\dfrac{{16}}{{35}}$
B) $\dfrac{{11}}{8}$
C) $\dfrac{{35}}{{16}}$
D) $\dfrac{8}{6}$

Answer 1

Hint: In order to find the sum of the infinite series, name the series $S$, then multiply both the sides by $\dfrac{1}{5}$, and then subtract the new obtained series from the first series taken. Use the concept of infinite geometric progression, solve the terms and get the results/correct option.

Complete step by step solution:
We are given with an infinite series: $1 + \dfrac{4}{5} + \dfrac{7}{{{5^2}}} + \dfrac{{10}}{{{5^3}}} + .....$
Naming the series as sum or $S$, so the series is:
$S = 1 + \dfrac{4}{5} + \dfrac{7}{{{5^2}}} + \dfrac{{10}}{{{5^3}}} + .....$ ……………..(1)
Multiplying both the sides of the equation by $\dfrac{1}{5}$, and we get:
$\dfrac{1}{5}S = \dfrac{1}{5}\left( {1 + \dfrac{4}{5} + \dfrac{7}{{{5^2}}} + \dfrac{{10}}{{{5^3}}} + .....} \right)$
Opening the braces on the right side:
$\dfrac{1}{5}S = \dfrac{1}{5} + \dfrac{4}{{{5^2}}} + \dfrac{7}{{{5^3}}} + \dfrac{{10}}{{{5^4}}} + .....$ ……………..(2)
Subtracting (2) from (1), and we get:
\[S - \dfrac{1}{5}S = \left( {1 + \dfrac{4}{5} + \dfrac{7}{{{5^2}}} + \dfrac{{10}}{{{5^3}}} + .....} \right) - \left( {\dfrac{1}{5} + \dfrac{4}{{{5^2}}} + \dfrac{7}{{{5^3}}} + \dfrac{{10}}{{{5^4}}} + .....} \right)\]
Opening the parenthesis on the right-side and solving it further, and similarly for the left-hand side:
\[
  S - \dfrac{1}{5}S = \left( {1 + \dfrac{4}{5} + \dfrac{7}{{{5^2}}} + \dfrac{{10}}{{{5^3}}} + .....} \right) - \left( {\dfrac{1}{5} + \dfrac{4}{{{5^2}}} + \dfrac{7}{{{5^3}}} + \dfrac{{10}}{{{5^4}}} + .....} \right) \\
   \Rightarrow \dfrac{{5 - 1}}{5}S = 1 + \dfrac{4}{5} + \dfrac{7}{{{5^2}}} + \dfrac{{10}}{{{5^3}}} - \dfrac{1}{5} - \dfrac{4}{{{5^2}}} - \dfrac{7}{{{5^3}}} - \dfrac{{10}}{{{5^4}}} + ...... \\
 \]
Solving similar types of constants, and we get:
\[
  \dfrac{{5 - 1}}{5}S = 1 + \dfrac{4}{5} + \dfrac{7}{{{5^2}}} + \dfrac{{10}}{{{5^3}}} - \dfrac{1}{5} - \dfrac{4}{{{5^2}}} - \dfrac{7}{{{5^3}}} - \dfrac{{10}}{{{5^4}}} + ...... \\
   \Rightarrow \dfrac{4}{5}S = 1 + \dfrac{3}{5} + \dfrac{3}{{{5^2}}} + \dfrac{3}{{{5^3}}} + ...... \\
 \]
Taking \[\dfrac{3}{5}\] common from the right-side except the first operand:
\[\dfrac{4}{5}S = 1 + \dfrac{3}{5}\left( {1 + \dfrac{1}{5} + \dfrac{1}{{{5^2}}} + .....} \right)\] ……..(3)
We get another geometric series inside the parenthesis \[1 + \dfrac{1}{5} + \dfrac{1}{{{5^2}}} + .....\]
Comparing the above obtained series with the general geometric series that is $a, ar, a{r^2}......,a{r^{n - 1}},.....$, we get first term $a = 1$, second term as $ar = \dfrac{1}{5}$ and so on.
Dividing first term by second term, we get:
$
  \dfrac{{\sec ond}}{{first}} = \dfrac{{ar}}{a} \\
  \dfrac{{ar}}{a} = \dfrac{{\dfrac{1}{5}}}{1} \\
  r = \dfrac{{\dfrac{1}{5}}}{1} = \dfrac{1}{5} \\
 $
Since, $\left| r \right| = \left| {\dfrac{1}{5}} \right| < 1$, so we use the formula ${S_\infty } = \dfrac{a}{{1 - r}}$, where $a$ is the first term and $r$ is the common ratio and ${S_\infty }$is the sum of the infinite series.
Substituting the value’s obtained in the formula ${S_\infty } = \dfrac{a}{{1 - r}}$, and we get:
$
  {S_\infty } = \dfrac{a}{{1 - r}} = \dfrac{1}{{1 - \left( {\dfrac{1}{5}} \right)}} \\
  {S_\infty } = \dfrac{1}{{1 - \dfrac{1}{5}}} = \dfrac{1}{{\dfrac{{5 - 1}}{5}}} = \dfrac{1}{{\dfrac{4}{5}}} = \dfrac{5}{4} \\
 $
Substituting this value in (3), we get:
\[\dfrac{4}{5}S = 1 + \dfrac{3}{5}\left( {\dfrac{5}{4}} \right)\]
Multiplying the braces value:
\[
  \dfrac{4}{5}S = 1 + \dfrac{3}{5}\left( {\dfrac{5}{4}} \right) \\
   \Rightarrow \dfrac{4}{5}S = 1 + \dfrac{{15}}{{20}} \\
 \]
Solving the right-hand side value:
\[
  \dfrac{4}{5}S = 1 + \dfrac{{15}}{{20}} \\
   \Rightarrow \dfrac{4}{5}S = \dfrac{{20 + 15}}{{20}} \\
   \Rightarrow \dfrac{4}{5}S = \dfrac{{35}}{{20}} \\
 \]
Multiplying both the terms by \[5\] and dividing by \[4\] in order to get \[S\]:
\[
   \Rightarrow \dfrac{4}{5}S = \dfrac{{35}}{{20}} \\
   \Rightarrow \dfrac{4}{5}S \times \dfrac{5}{4} = \dfrac{{35}}{{20}} \times \dfrac{5}{4} \\
   \Rightarrow S = \dfrac{{35}}{{16}} \\
 \]
Which matches with the third option.

Therefore, the sum of infinity of the series $1 + \dfrac{4}{5} + \dfrac{7}{{{5^2}}} + \dfrac{{10}}{{{5^3}}} + .....$ is Option C : $\dfrac{{35}}{{16}}$.

Additional Information:
A finite series is a series whose last term is known, in other words the number of terms are known to us and is represented by $a, ar, a{r^2}......,a{r^n}$, where $a$ is the first term and $r$ is the common ratio.

An infinite geometric series sum is the sum of an infinite geometric sequence. And thus, we know that infinite means not having any last term, so infinite series does not have any last term, and the series is represented by $a, ar, a{r^2}......,a{r^{n - 1}},.....$, where $a$ is the first term and $r$ is the common ratio.

Note:
1) Always preferred to go step by step in order to reduce the chances of mistakes.
2) There is a huge difference between infinite and finite geometric series, do not consider them the same.