Question

The sum of $n$ terms of the series 1.4+3.04+5.004+7.0004+..... is
A. ${n^2} + \dfrac{4}{9}\left( {1 + \dfrac{1}{{{{10}^n}}}} \right)$
B. ${n^2} + \dfrac{4}{9}\left( {1 - \dfrac{1}{{{{10}^n}}}} \right)$
C. $n + \dfrac{4}{9}\left( {1 - \dfrac{1}{{{{10}^n}}}} \right)$
D. None of these

Answer 1

We will rewrite the given series as 1+3+5+7....+0.4+0.04+0.004+0.0004+…. Then we will calculate the sum of series $\;{S_1} = 1 + 3 + 5 + 7...$ which is an AP and the sum of $n$ terms of an AP is given by $\dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$ where $a$ is the first term and $d$ is the common difference. Then we will find the sum of the series ${S_2}{\text{ = }}0.4 + 0.04 + 0.004 + ....$ where the sum of the GP is given by $\dfrac{{a\left( {1 - {r^n}} \right)}}{{1 - r}}$ where $1 > r$.

Complete step-by-step answer:
We have to calculate the sum of n terms of the series 1.4+3.04+5.004+7.0004+..... .
We will rewrite the series by separating the decimal part of the number.
1+0.4+3+0.04+5+0.004+7+0.0004+....
On rearranging the terms of the above series we get
1+3+5+7....+0.4+0.04+0.004+0.0004+...
The first part of the above series forms an AP whereas the other part of the AP forms a GP.
The sum of $n$ terms an AP is given as $\dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$ where $a$ is the first term and $d$ is the common difference.
We will first find the sum $\;{S_1} = 1 + 3 + 5 + 7...$
Here the first term is 1 and common difference is $3 - 1 = 2$
$\Rightarrow {S_1} = \dfrac{n}{2}\left( {2\left( 1 \right) + \left( {n - 1} \right)2} \right) \\\ \Rightarrow {S_1} = n\left( {1 + \left( {n - 1} \right)} \right) \\\ {S_1} = {n^2} \\\$
Next we will calculate the sum ${S_2}{\text{ = }}0.4 + 0.04 + 0.004 + ....$
Here the first term is 0.4 and the common ratio is $\dfrac{{0.04}}{{0.4}} = 0.1$
The sum of $n$ terms of GP is given by $\dfrac{{a\left( {1 - {r^n}} \right)}}{{1 - r}}$ where $1 > r$.
On substituting the values we get,
$\Rightarrow {S_2} = \dfrac{{0.4\left( {1 - {{\left( {0.1} \right)}^n}} \right)}}{{1 - \left( {0.1} \right)}} \\\ \Rightarrow {S_2} = \dfrac{{4\left( {1 - {{\left( {0.1} \right)}^n}} \right)}}{9} \\\$
The required sum will be
$S = {S_1} + {S_2}$
$\Rightarrow S = {n^2} + \dfrac{{4\left( {1 - {{\left( {0.1} \right)}^n}} \right)}}{9} \\\ \Rightarrow S = {n^2} + \dfrac{4}{9}\left( {1 - \dfrac{1}{{{{10}^n}}}} \right) \\\$
Hence, option B is correct.

Note: The GP is of the form $a,ar,{\text{ }}a{r^2},.....$ where $a$ is the first term and $r$ is the common ratio. The sum of GP where $1 > r$, is $\dfrac{{a\left( {1 - {r^n}} \right)}}{{1 - r}}$ And when $r > 1$ the sum of the series of GP is $\dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}$