Question

The sum of first twenty terms of the series $1 + \dfrac{3}{2} + \dfrac{7}{4} + \dfrac{{15}}{8} + \dfrac{{31}}{{16}} + .......$ , is
$a)\,38 + \dfrac{1}{{{{20}^{20}}}} \\\ b)\,39 + \dfrac{1}{{{{20}^{19}}}} \\\ c)\,39 + \dfrac{1}{{{{20}^{20}}}} \\\ d)\,38 + \dfrac{1}{{{{20}^{19}}}}$

Answer 1

In this question we are given a series and we have to find the sum of the first twenty terms of the series. In order to calculate the sum we will find the ${n^{th}}$ term of the given series. And using ${n^{th}}$ term we can easily find the sum of the first twenty terms.

Complete step-by-step answer:
We are given a series:
$1 + \dfrac{3}{2} + \dfrac{7}{4} + \dfrac{{15}}{8} + \dfrac{{31}}{{16}} + .......\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \left( 1 \right)$
And we have to find the sum of first $20$ terms
To solve the question we will take the series and find its ${n^{th}}$ term
So, we can write (1) as,
$\;1 + \dfrac{{4 - 1}}{{{2^1}}} + \dfrac{{8 - 1}}{{{2^2}}} + \dfrac{{16 - 1}}{{{2^3}}} + \dfrac{{32 - 1}}{{{2^4}}} + ........$
Because $2 = {2^1},\,4 = {2^2}$ and so on,
Also, $\dfrac{{{2^1} - 1}}{{{2^0}}} + \dfrac{{{2^2} - 1}}{{{2^1}}} + \dfrac{{{2^3} - 1}}{{{2^2}}} + \dfrac{{{2^4} - 1}}{{{2^3}}} + \dfrac{{{2^5} - 1}}{{{2^4}}} + ........\,\,\,\,\,\,\,\, \to \left( 3 \right)$
So clearly from (3) we can say that ${n^{th}}$ term for the series (1) is
${T_n} = \dfrac{{{2^n}}}{{{2^{n - 1}}}} - \dfrac{1}{{{2^{n - 1}}}}\,\,\,\,\,\,\,\,\,\, \to \left( 4 \right)$
According to the question we have to find the sum of first twenty terms
Let S be the sum of the first twenty terms. So,

$$\Rightarrow S = \sum\limits_{n = 1}^{20} {{T_n}} \\\ \Rightarrow S = \sum\limits_{n = 1}^{20} {\dfrac{{{2^n}}}{{{2^{n - 1}}}} - \dfrac{1}{{{2^{n - 1}}}}\,} \\\$$

Now, in the expression $\dfrac{{{2^n}}}{{{2^{n - 1}}}}$, base is same so powers will be subtracted, then

$$\Rightarrow S = \sum\limits_{n = 1}^{20} {{2^{n - n + 1}} - \dfrac{1}{{{2^{n - 1}}}}\,} \\\ \Rightarrow S = \sum\limits_{n = 1}^{20} {2 - \dfrac{1}{{{2^{n - 1}}}}\,} \,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \left( 5 \right) \\\$$

Now we know that $\sum\limits_{i = 1}^n {{a_i} + {b_i}} = \sum\limits_{i = 1}^n {{a_i}} + \sum\limits_{i = 1}^n {{b_i}}$
So using this in (5) where ${a_i} = 2\,\,\,,\,\,\,{b_i} = \dfrac{1}{{{2^{n - 1}}}}$
So
$S = \sum\limits_{n = 1}^{20} {2\,} - \sum\limits_{n = 1}^{20} {\dfrac{1}{{{2^{n - 1}}}}\,}$
Now we know that $\sum\limits_{n = 1}^m a = am$ where $a$ is any constant, so using it here in first summation where $a = 2$ and $m = 20$, we get
$S = 2(20) - \sum\limits_{n = 1}^{20} {\dfrac{1}{{{2^{n - 1}}}}\,} \,\,\,\,\,\,\,\,\,\,\,\,\, \to \left( 6 \right)$
Now consider
$\sum\limits_{n = 1}^{20} {\dfrac{1}{{{2^{n - 1}}}}\,} = \dfrac{1}{{{2^0}}} + \dfrac{1}{{{2^1}}} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + ......... + \dfrac{1}{{{2^{19}}}}$
It can be clearly seen that a geometric progression with common ratio $\dfrac{1}{2}$ is formed here.
So, the sum of first $n$ terms of this G.P. with common ratio $r$ and first term $a$ is
${S_n} = \dfrac{{a\left( {1 - {r^n}} \right)}}{{1 - r}}\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \left( 7 \right)$
So now using (7) for $\sum\limits_{n = 1}^{20} {\dfrac{1}{{{2^{n - 1}}}}\,} \,\,$ where $a = 1,r = \dfrac{1}{2},n = 20$
So we get

$$\Rightarrow \sum\limits_{n = 1}^{20} {\dfrac{1}{{{2^{n - 1}}}}\,} \,\, = \dfrac{{1\left( {1 - {{\left( {\dfrac{1}{2}} \right)}^{20}}} \right)}}{{1 - \dfrac{1}{2}}} = \dfrac{{1 - \dfrac{1}{{{2^{20}}}}}}{{\dfrac{1}{2}}} = 2\left( {1 - \dfrac{1}{{{2^{20}}}}} \right) \\\ \Rightarrow \sum\limits_{n = 1}^{20} {\dfrac{1}{{{2^{n - 1}}}}\,} \,\, = 2\left( {1 - \dfrac{1}{{{2^{20}}}}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \left( 8 \right) \\\$$

Now substituting this value in (6) we get
$\Rightarrow S = 40 - \left( {2 - \dfrac{2}{{{2^{20}}}}} \right) \\\ \Rightarrow S = 38 + \dfrac{2}{{{2^{20}}}} \\\ \Rightarrow S = 38 + \dfrac{1}{{{2^{19}}}}$
So the sum of first twenty terms of given series is $38 + \dfrac{1}{{{2^{19}}}}$

So option D is the correct answer.

Note: The tricky part of the question is to find the ${n^{th}}$ term for the series.
And here to find ${n^{th}}$ term for the series we will proceed as firstly,
We can clearly see that the given series is neither A.P. nor G.P.
So now we will try to observe a pattern.
Here, looking at denominators of each term we can see that denominators are powers of two.
And numerators are one less than the power of two.
So, by observing these types of patterns we can solve this question very easily.