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# The sum of the series $\dfrac{5}{13}+\dfrac{55}{{{13}^{2}}}+\dfrac{555}{{{13}^{3}}}+.....\infty$ is(a) $\dfrac{65}{36}$(b) $\dfrac{65}{32}$(c) $\dfrac{25}{36}$(d) None of these

Last updated date: 12th Jul 2024
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Hint: In this series, first of all, we will multiply $\dfrac{1}{13}$ on both sides and then subtract this series from the original series. Then G.P would be obtained. Use the formula for the sum of G.P that is ${{S}_{\infty }}=\dfrac{a}{1-r}$ to get the desired sum.

Here, we have to find the sum of the series $\dfrac{5}{13}+\dfrac{55}{{{13}^{2}}}+\dfrac{555}{{{13}^{3}}}+.....\infty$.
Let us consider the series given in the question.
$S=\dfrac{5}{13}+\dfrac{55}{{{13}^{2}}}+\dfrac{555}{{{13}^{3}}}+.....\infty ....\left( i \right)$
By multiplying both the sides of the above expression by $\dfrac{1}{13}$, we get,
$\dfrac{S}{13}=\dfrac{5}{{{\left( 13 \right)}^{2}}}+\dfrac{55}{{{13}^{3}}}+\dfrac{555}{{{13}^{4}}}+.....\infty ....\left( ii \right)$
Now, by subtracting equation (ii) from equation (i), we get,
$S-\dfrac{S}{13}=\dfrac{5}{13}+\left( \dfrac{55}{{{13}^{2}}}-\dfrac{5}{{{\left( 13 \right)}^{2}}} \right)+\left( \dfrac{555}{{{13}^{3}}}-\dfrac{55}{{{13}^{3}}} \right)+\left( \dfrac{5555}{{{13}^{4}}}-\dfrac{555}{{{13}^{4}}} \right)+.....\infty$
By simplifying the above equation, we get,
$\dfrac{12S}{13}=\dfrac{5}{13}+\left( \dfrac{50}{{{\left( 13 \right)}^{2}}} \right)+\left( \dfrac{500}{{{13}^{3}}} \right)+\left( \dfrac{5000}{{{13}^{4}}} \right)+.....\infty$
Now by dividing both the sides of the above equation by 5, we get,
$\dfrac{1}{5}.\dfrac{12S}{13}=\dfrac{1}{13}+\dfrac{10}{{{\left( 13 \right)}^{2}}}+\dfrac{100}{{{13}^{3}}}+\dfrac{1000}{{{13}^{4}}}+.....\infty$
Or we can also write the above equation as,
$\dfrac{12S}{65}=\dfrac{1}{13}+\dfrac{10}{{{\left( 13 \right)}^{2}}}+\dfrac{{{\left( 10 \right)}^{2}}}{{{13}^{3}}}+\dfrac{{{\left( 10 \right)}^{3}}}{{{13}^{4}}}+.....\infty ....\left( iii \right)$
Now, we know the terms of the G.P are of the form $a,ar,a{{r}^{2}},a{{r}^{3}}.....$
By comparing the terms of RHS of the equation (iii), with the general terms of the G.P, we get, $a=\dfrac{1}{13}$ and $r=\dfrac{10}{13}$.
We know that the sum of the terms of the infinite G.P $=\dfrac{a}{1-r}$. By using this in the RHS of equation (iii), we get,
$\dfrac{12S}{65}=\dfrac{\dfrac{1}{13}}{1-\dfrac{10}{13}}$
By simplifying the above equation, we get,
$\dfrac{12S}{65}=\dfrac{\dfrac{1}{13}}{\dfrac{13-10}{13}}$
Or, $\dfrac{12S}{65}=\dfrac{1}{3}$
By cross multiplying the above equation, we get,
$S=\dfrac{65}{3\times 12}=\dfrac{65}{36}$
So, we get the sum of the series as $\dfrac{65}{36}$.
Hence, option (a) is the right answer.

Note: In these types of questions, always try to transform the series into geometric progression by multiplying a suitable number with it. This approach should be there in mind. Also, take special care whether the given series is finite or infinite and use the formula for the sum of the finite series as ${{S}_{n}}=\dfrac{a\left( 1-{{r}^{n}} \right)}{\left( 1-r \right)}$ and for the sum of the infinite series as ${{S}_{\infty }}=\dfrac{a}{\left( 1-r \right)}$.