 # The sum of $1 + \dfrac{2}{5} + \dfrac{3}{{{5^2}}} + \dfrac{4}{{{5^3}}} + ....$ up to $n$ terms isA) $\left( {\dfrac{{25}}{{16}}} \right) - \dfrac{{\left( {4n + 5} \right)}}{{\left( {{{16.5}^{n - 1}}} \right)}}$B) $\left( {\dfrac{3}{4}} \right) - \dfrac{{\left( {2n + 5} \right)}}{{\left( {{{16.5}^{n - 1}}} \right)}}$ C) $\left( {\dfrac{3}{7}} \right) - \dfrac{{\left( {3n + 5} \right)}}{{\left( {{{16.5}^{n - 1}}} \right)}}$D) $\left( {\dfrac{1}{2}} \right) - \dfrac{{\left( {5n + 1} \right)}}{{\left( {{{3.5}^{n + 2}}} \right)}}$ Verified
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Hint: We are to first assume the given sum as any arbitrary number as $S$. Then, depending on the question we are to multiply any scalar quantity with $S$ such to make it as $kS$ and subtract it from $S$ in such a way that the first number of $kS$ corresponds to the second number of $S$. Then, we significantly operate the sum obtained in order to get to a suitable answer.

Complete step by step solution:
Let $S = 1 + \dfrac{2}{5} + \dfrac{3}{{{5^2}}} + \dfrac{4}{{{5^3}}} + .... + \dfrac{n}{{{5^{n - 1}}}} - - - \left( 1 \right)$
Now, multiplying both sides with $\dfrac{1}{5}$, we get,
$\dfrac{S}{5} = \dfrac{1}{5} + \dfrac{2}{{{5^2}}} + \dfrac{3}{{{5^3}}} + \dfrac{4}{{{5^4}}} + .... + \dfrac{n}{{{5^n}}} - - - \left( 2 \right)$
Now, $\left( 1 \right) - \left( 2 \right)$, gives us,
$\dfrac{{4S}}{5} = 1 + \dfrac{1}{5} + \dfrac{1}{{{5^2}}} + \dfrac{1}{{{5^3}}} + ... - \dfrac{n}{{{5^n}}}$
Therefore, we can see that, this is clearly a GP with first term $1$ and common ratio $\dfrac{1}{5}$ excluding the last term $\dfrac{n}{{{5^n}}}$.
Therefore, we know, the formula for sum of $n$ terms of a GP is $\dfrac{{a(1 - {r^n})}}{{(1 - r)}}$, where $a$ is the first term of the GP and $r$ is the common ratio of the GP.
Using this formula in the above equation, we get,
$\Rightarrow \dfrac{{4S}}{5} = \left[ {1 + \dfrac{1}{5} + \dfrac{1}{{{5^2}}} + \dfrac{1}{{{5^3}}} + ...} \right] - \dfrac{n}{{{5^n}}}$
$\Rightarrow \dfrac{{4S}}{5} = \dfrac{{1\left( {1 - {{\left( {\dfrac{1}{5}} \right)}^n}} \right)}}{{\left( {1 - \dfrac{1}{5}} \right)}} - \dfrac{n}{{{5^n}}}$
$\Rightarrow \dfrac{{4S}}{5} = \dfrac{{\left( {1 - \dfrac{1}{{{5^n}}}} \right)}}{{\dfrac{4}{5}}} - \dfrac{n}{{{5^n}}}$
Now, simplifying, we get,
$\Rightarrow \dfrac{{4S}}{5} = \dfrac{{{5^n} - 1}}{{{{4.5}^{n - 1}}}} - \dfrac{n}{{{5^n}}}$
Taking the LCM on right hand side of the equation and simplifying it, we get,
$\Rightarrow \dfrac{{4S}}{5} = \dfrac{{{5^{n + 1}} - 5 - 4n}}{{{{4.5}^n}}}$
Multiplying both sides by $\dfrac{5}{4}$, we get,
$\Rightarrow S = \left( {\dfrac{{{5^{n + 1}} - 4n - 5}}{{{{4.5}^n}}}} \right).\dfrac{5}{4}$
Now, simplifying the terms, we get,
$\Rightarrow S = \dfrac{{{5^{n + 1}} - 4n - 5}}{{{{16.5}^{n - 1}}}}$
$\Rightarrow S = \dfrac{{{5^{n + 1}} - (4n + 5)}}{{{{16.5}^{n - 1}}}}$
Dividing the numerator by the denominator, we get,
$\Rightarrow S = \dfrac{{{5^{n + 1}}}}{{{{16.5}^{n - 1}}}} - \dfrac{{4n + 5}}{{{{16.5}^{n - 1}}}}$
On simplifying, we get,
$\Rightarrow S = \dfrac{{{5^2}}}{{16}} - \dfrac{{4n + 5}}{{{{16.5}^{n - 1}}}}$
$\Rightarrow S = \left( {\dfrac{{25}}{{16}}} \right) - \left( {\dfrac{{4n + 5}}{{{{16.5}^{n - 1}}}}} \right)$
Therefore, the sum of the series is $\left( {\dfrac{{25}}{{16}}} \right) - \left( {\dfrac{{4n + 5}}{{{{16.5}^{n - 1}}}}} \right)$, the correct answer is 1.

Note:
In these questions, we are to just obtain a sequence in which we can easily add the terms, either in a sequence, for which we know the sum of the terms up to $n$ terms, like the sum of first $n$ natural numbers or the sum of squares of first $n$ natural numbers. Or in the form of AP, GP or HP, as we can find the sum of $n$ terms of these sequences.