Question

The sum of the series $\sum\limits_{r = 0}^n {{{( - 1)}^r}^n} {C_r}\left( {\dfrac{1}{{{2^r}}} + \dfrac{{{3^r}}}{{{2^{2r}}}} + \dfrac{{{7^r}}}{{{2^{2r}}}} + \dfrac{{{{15}^r}}}{{{2^{2r}}}} + .......{\text{m terms}}} \right)$ is
A) $\dfrac{{{2^{mn}} - 1}}{{{2^{mn}}({2^n} - 1)}}$
B) $\dfrac{{{2^{mn}} - 1}}{{{2^n} - 1}}$
C) $\dfrac{{{2^{mn}} + 1}}{{{2^n} + 1}}$
D) None of these

Answer 1

The sum of the series is given in the question. We have to simplify and find that the given particular option is which one is the correct option that is equal to the sum of the given series. First, we will expand the given series. Then, we will simplify the terms of the given series by using some series relation. After that, we will apply the formula of combination. We will get the final solution.

Formula used: $\sum\limits_{r = 0}^n {{{( - 1)}^r}^n} {C_r}{x^r} = {(1 - x)^n}$.

Complete step-by-step solution:
It is given that: the given series is $\sum\limits_{r = 0}^n {{{( - 1)}^r}^n} {C_r}\left( {\dfrac{1}{{{2^r}}} + \dfrac{{{3^r}}}{{{2^{2r}}}} + \dfrac{{{7^r}}}{{{2^{2r}}}} + \dfrac{{{{15}^r}}}{{{2^{2r}}}} + .......{\text{m terms}}} \right)$
We have to find the sum of the series.
$\sum\limits_{r = 0}^n {{{( - 1)}^r}^n} {C_r}\left( {\dfrac{1}{{{2^r}}} + \dfrac{{{3^r}}}{{{2^{2r}}}} + \dfrac{{{7^r}}}{{{2^{2r}}}} + \dfrac{{{{15}^r}}}{{{2^{2r}}}} + .......{\text{m terms}}} \right)$
Simplifying we get,
$\Rightarrow \sum\limits_{r = 0}^n {{{( - 1)}^r}} {\left( {\dfrac{1}{2}} \right)^r} + \sum\limits_{r = 0}^n {{{( - 1)}^r}^n{C_r}} {\left( {\dfrac{3}{4}} \right)^r} + \sum\limits_{r = 0}^n {{{( - 1)}^r}^n{C_r}} {\left( {\dfrac{7}{8}} \right)^r} +$… up to m terms
We know that,
$\sum\limits_{r = 0}^n {{{( - 1)}^r}^n} {C_r}{x^r} = {(1 - x)^n}$
Using this rule, we get,
$\Rightarrow {\left( {1 - \dfrac{1}{2}} \right)^n} + {\left( {1 - \dfrac{3}{4}} \right)^n} + {\left( {1 - \dfrac{7}{8}} \right)^n} +$… up to m terms
Simplifying we get,
$\Rightarrow {\left( {\dfrac{1}{2}} \right)^n} + {\left( {\dfrac{1}{4}} \right)^n} + {\left( {\dfrac{1}{8}} \right)^n} +$… up to m terms
It is geometric series with initial term as $= {\left( {\dfrac{1}{2}} \right)^n}$ and the common ratio is ${\left( {\dfrac{1}{2}} \right)^n}$.
We know that, the general form of sum of geometric series is $= \dfrac{{a(1 - {r^n})}}{{1 - r}}$when $r < 1$.
Simplifying again we get,
$\Rightarrow {\left( {\dfrac{1}{2}} \right)^n}\left[ {\dfrac{{1 - {{\left( {\dfrac{1}{{{2^n}}}} \right)}^m}}}{{1 - \dfrac{1}{{{2^n}}}}}} \right]$
Simplifying again we get,
$\Rightarrow \dfrac{{{2^{mn}} - 1}}{{{2^{mn}}({2^n} - 1)}}$

$\therefore$ The correct option is A.

Note: A series can be highly generalized as the sum of all the terms in a sequence. However, there has to be a definite relationship between all the terms of the sequence.
The general form of a series is: $\sum\limits_{i = 0}^n {{a_i}}$
Which can be written as: ${a_0} + {a_1} + {a_3} + ... + {a_n}$.
There are two types of series:
Finite series: Here, the number of elements of the series is finite.
The general form of a finite series is: $\sum\limits_{i = 0}^n {{a_i}}$
Which can be written as: ${a_0} + {a_1} + {a_3} + ... + {a_n}$.