Question

Find the sum of an infinite series \[1 + \dfrac{4}{5} + \dfrac{7}{{{5^2}}} + \dfrac{{10}}{{{5^3}}} + .....\infty \].
A. \[\dfrac{7}{{16}}\]
B. \[\dfrac{5}{{16}}\]
C. \[\dfrac{{104}}{{64}}\]
D. \[\dfrac{{35}}{{16}}\]

Answer 1

Hint In the given question, the infinite series is given. We will divide the series by 5, and then subtract the new series from the original series. The terms of the new series are in geometric progression. So, by using the formula of the sum of infinite terms in geometric progression, we will find the value of the series.

Formula used
The sum of infinite terms in GP: \[{S_\infty } = \dfrac{a}{{\left( {1 - r} \right)}}\] , \[\left| r \right| < 1\]
where \[a\] is the first term and \[r\] is the common ratio.

Complete step by step solution:
The given series is,
    \[S = 1 + \dfrac{4}{5} + \dfrac{7}{{{5^2}}} + \dfrac{{10}}{{{5^3}}} + .....\infty \] \[.......equation\left( 1 \right)\]
Divide the above series by 5.
    \[\dfrac{S}{5} = \dfrac{1}{5} + \dfrac{4}{{{5^2}}} + \dfrac{7}{{{5^3}}} + \dfrac{{10}}{{{5^4}}} + .....\infty \] \[.......equation\left( 2 \right)\]
Subtract equation \[\left( 2 \right)\] from equation \[\left( 1 \right)\].
\[S - \dfrac{S}{5} = \left( {1 + \dfrac{4}{5} + \dfrac{7}{{{5^2}}} + \dfrac{{10}}{{{5^3}}} + .....\infty } \right) - \left( {\dfrac{1}{5} + \dfrac{4}{{{5^2}}} + \dfrac{7}{{{5^3}}} + \dfrac{{10}}{{{5^4}}} + .....\infty } \right)\]
Simplify the above equation.
Subtract the terms of second bracket from the terms of first brackets with same denominator.
\[\dfrac{{4S}}{5} = 1 + \dfrac{{4 - 1}}{5} + \dfrac{{7 - 4}}{{{5^2}}} + \dfrac{{10 - 7}}{{{5^3}}} + .....\infty \]
\[ \Rightarrow \]\[\dfrac{{4S}}{5} = 1 + \dfrac{3}{5} + \dfrac{3}{{{5^2}}} + \dfrac{3}{{{5^3}}} + .....\infty \]
\[ \Rightarrow \]\[\dfrac{{4S}}{5} = 1 + 3\left[ {\dfrac{1}{5} + \dfrac{1}{{{5^2}}} + \dfrac{1}{{{5^3}}}.....\infty } \right]\]
The terms in the square bracket are in geometric progression where the first term is \[a = \dfrac{1}{5}\] and the common ratio is \[r = \dfrac{{\dfrac{1}{{{5^2}}}}}{{\dfrac{1}{5}}} = \dfrac{1}{5}\].
Apply the formula of the sum of infinite terms in geometric progression \[{S_\infty } = \dfrac{a}{{\left( {1 - r} \right)}}\] in the square bracket.
\[\dfrac{{4S}}{5} = 1 + 3\left[ {\dfrac{{\dfrac{1}{5}}}{{\left( {1 - \dfrac{1}{5}} \right)}}} \right]\]
\[ \Rightarrow \]\[\dfrac{{4S}}{5} = 1 + 3\left[ {\dfrac{{\dfrac{1}{5}}}{{\dfrac{4}{5}}}} \right]\]
\[ \Rightarrow \]\[\dfrac{{4S}}{5} = 1 + 3\left[ {\dfrac{1}{4}} \right]\]
\[ \Rightarrow \]\[\dfrac{{4S}}{5} = 1 + \dfrac{3}{4}\]
\[ \Rightarrow \]\[\dfrac{{4S}}{5} = \dfrac{7}{4}\]
\[ \Rightarrow \]\[S = \dfrac{7}{4} \times \dfrac{5}{4}\]
\[ \Rightarrow \]\[S = \dfrac{{35}}{{16}}\]

Hence the correct option is option D.

Note: Students are often confused with the value of \[\dfrac{{\left( {\dfrac{a}{b}} \right)}}{{\left( {\dfrac{c}{d}} \right)}}\] that whether \[\left( {\dfrac{{ac}}{{bd}}} \right)\] or \[\left( {\dfrac{{ad}}{{bc}}} \right)\]. But the correct formula is \[\dfrac{{\left( {\dfrac{a}{b}} \right)}}{{\left( {\dfrac{c}{d}} \right)}} = \dfrac{{ad}}{{bc}}\].