If \[{S_n} = \sum\limits_{r = 1}^n {\dfrac{{1 + 2 + {2^2} + \ldots \ldots {\rm{upto\, }}r\,{\rm{ terms}}}}{{{2^r}}}} \], then \[{S_n}\] is equal to
(a) \[2n - \left( {n + 1} \right)\]
(b) \[1 - \dfrac{1}{{{2^n}}}\]
(c) \[n - 1 + \dfrac{1}{{{2^n}}}\]
(d) \[1 + \dfrac{1}{{{2^n}}}\]

Hint: Here, we need to find the value of \[{S_n}\]. We will use the formula for sum of \[n\] terms of a geometric progression to simplify the given equation, and thus, find which of the given options is the correct value of \[{S_n}\]. A geometric progression is a series of numbers in which each successive number is the product of the previous number and a fixed ratio.

Formula Used:
We will use the formula of the sum of \[n\] terms of a G.P. is given by the formula \[\dfrac{{a\left( {{R^n} - 1} \right)}}{{R - 1}}\], where \[a\] is the first term, \[R\] is the common ratio, and \[n\] is the number of terms of the G.P.

Complete step-by-step answer:
We will simplify the given equation \[{S_n} = \sum\limits_{r = 1}^n {\dfrac{{1 + 2 + {2^2} + \ldots \ldots {\rm{upto\, }}r\,{\rm{ terms}}}}{{{2^r}}}} \] to find the value of \[{S_n}\].
The numerator of the expression is \[1 + 2 + {2^2} + \ldots \ldots {\rm{upto\, }}r\,{\rm{ terms}}\].
Each successive term is equal to the product of 2, and the previous term.
Thus, we can observe that this series forms a geometric progression where the first term is 1, and the common ratio is 2.
Now, we will find the sum of this G.P. using \[\dfrac{{a\left( {{R^n} - 1} \right)}}{{R - 1}}\].
The first term of the G.P. is 1.
Thus, we get
\[a = 1\]
The common ratio of the G.P. is 2.
Thus, we get
\[R = 2\]
The number of terms in the G.P. is \[r\].
Thus, we get
\[n = r\]
Substituting \[n = r\], \[R = 2\], and \[a = 1\] in the formula for sum of terms of a G.P., \[\dfrac{{a\left( {{R^n} - 1} \right)}}{{R - 1}}\], we get
\[ \Rightarrow 1 + 2 + {2^2} + \ldots \ldots {\rm{upto\, }}r\,{\rm{ terms}} = \dfrac{{1\left( {{2^r} - 1} \right)}}{{2 - 1}}\]
Simplifying the expression, we get
\[\begin{array}{l} \Rightarrow 1 + 2 + {2^2} + \ldots \ldots {\rm{upto\, }}r\,{\rm{ terms}} = \dfrac{{{2^r} - 1}}{1}\\ \Rightarrow 1 + 2 + {2^2} + \ldots \ldots {\rm{upto\, }}r\,{\rm{ terms}} = {2^r} - 1\end{array}\]
Substituting \[1 + 2 + {2^2} + \ldots \ldots {\rm{upto\, }}r\,{\rm{ terms}} = {2^r} - 1\] in the equation for the sum \[{S_n} = \sum\limits_{r = 1}^n {\dfrac{{1 + 2 + {2^2} + \ldots \ldots {\rm{upto\, }}r\,{\rm{ terms}}}}{{{2^r}}}} \], we get
\[ \Rightarrow {S_n} = \sum\limits_{r = 1}^n {\dfrac{{{2^r} - 1}}{{{2^r}}}} \]
Rewriting the expression by splitting the L.C.M., we get
\[ \Rightarrow {S_n} = \sum\limits_{r = 1}^n {\left( {\dfrac{{{2^r}}}{{{2^r}}} - \dfrac{1}{{{2^r}}}} \right)} \]
Simplifying the expression, we get
\[ \Rightarrow {S_n} = \sum\limits_{r = 1}^n {\left( {1 - \dfrac{1}{{{2^r}}}} \right)} \]
We know that an expression of the form \[\sum\limits_{x = 1}^n {\left[ {f\left( x \right) - g\left( x \right)} \right]} \] can be written as \[\sum\limits_{x = 1}^n {f\left( x \right)} - \sum\limits_{x = 1}^n {g\left( x \right)} \].
Therefore, rewriting the expression, we get
\[ \Rightarrow {S_n} = \sum\limits_{r = 1}^n {\left( 1 \right)} - \sum\limits_{r = 1}^n {\left( {\dfrac{1}{{{2^r}}}} \right)} \]
Expanding both the sums, we get
\[ \Rightarrow {S_n} = \left( {1 + 1 + 1 + \ldots \ldots {\rm{upto\, }}n\,{\rm{ terms}}} \right) - \left( {\dfrac{1}{{{2^1}}} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + \ldots \ldots {\rm{upto\, }}n\,{\rm{ terms}}} \right)\]
In the first time, 1 is added \[n\] times. This repeated addition can be denoted as the product of 1 and \[n\].
Therefore, we get
\[\begin{array}{l} \Rightarrow {S_n} = \left( {1 \times n} \right) - \left( {\dfrac{1}{{{2^1}}} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + \ldots \ldots {\rm{upto\, }}n\,{\rm{ terms}}} \right)\\
\Rightarrow {S_n} = n - \left( {\dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + \ldots \ldots {\rm{upto\, }}n\,{\rm{ terms}}} \right)\end{array}\]
The second sum in the expression is \[\dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + \ldots \ldots {\rm{upto \,}}n\,{\rm{ terms}}\].
Each successive term is equal to the product of \[\dfrac{1}{2}\], and the previous term.
Thus, we can observe that this series forms a geometric progression where the first term is \[\dfrac{1}{2}\], and the common ratio is \[\dfrac{1}{2}\].
Now, we will find the sum of this G.P.
The first term of the G.P. is \[\dfrac{1}{2}\].
Thus, we get
\[a = \dfrac{1}{2}\]
The common ratio of the G.P. is \[\dfrac{1}{2}\].
Thus, we get
\[R = \dfrac{1}{2}\]
The number of terms in the G.P. is \[n\].
Substituting \[R = \dfrac{1}{2}\] and \[a = \dfrac{1}{2}\] in the formula for sum of terms of a G.P., we get
\[ \Rightarrow \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + \ldots \ldots {\rm{upto\, }}n\,{\rm{ terms}} = \dfrac{{\dfrac{1}{2}\left[ {{{\left( {\dfrac{1}{2}} \right)}^n} - 1} \right]}}{{\dfrac{1}{2} - 1}}\]
Simplifying the expression, we get
\[\begin{array}{l} \Rightarrow \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + \ldots \ldots {\rm{upto\, }}n\,{\rm{ terms}} = \dfrac{{\dfrac{1}{2}\left[ {{{\left( {\dfrac{1}{2}} \right)}^n} - 1} \right]}}{{ - \dfrac{1}{2}}}\\ \Rightarrow \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + \ldots \ldots {\rm{upto\, }}n\,{\rm{ terms}} = - \left[ {{{\left( {\dfrac{1}{2}} \right)}^n} - 1} \right]\\ \Rightarrow \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + \ldots \ldots {\rm{upto\, }}n\,{\rm{ terms}} = 1 - {\left( {\dfrac{1}{2}} \right)^n}\end{array}\]
An expression of the form \[{\left( {\dfrac{a}{b}} \right)^m}\] can be written as \[\dfrac{{{a^m}}}{{{b^m}}}\]. This is a rule of exponent.
Using the rule of exponent to rewrite \[{\left( {\dfrac{1}{2}} \right)^n}\], we get
\[ \Rightarrow \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + \ldots \ldots {\rm{upto\, }}n\,{\rm{ terms}} = 1 - \dfrac{{{1^n}}}{{{2^n}}}\]
The number 1 raised to any real exponent is equal to 1.
Therefore, we get
\[ \Rightarrow \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + \ldots \ldots {\rm{upto\, }}n\,{\rm{ terms}} = 1 - \dfrac{1}{{{2^n}}}\]
Finally, substituting \[\dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + \ldots \ldots {\rm{upto\, }}n\,{\rm{ terms}} = 1 - \dfrac{1}{{{2^n}}}\] in the equation \[{S_n} = n - \left( {\dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + \ldots \ldots {\rm{upto\, }}n\,{\rm{ terms}}} \right)\], we get
\[ \Rightarrow {S_n} = n - \left( {1 - \dfrac{1}{{{2^n}}}} \right)\]
Rewriting the expression, we get
\[\therefore {{S}_{n}}=n-1+\dfrac{1}{{{2}^{n}}}\]
Therefore, the sum \[{S_n}\] is equal to \[n - 1 + \dfrac{1}{{{2^n}}}\].
Thus, the correct option is option (c).

Note: In geometric progression, consecutive numbers differ by a fixed ratio and this fixed ratio is called the common ratio.
A common mistake is to use the formula for the sum of a geometric progression with infinite terms, that is \[\dfrac{a}{{1 - R}}\]. This is incorrect since it is given that the sum extends up to \[r\] or \[n\] terms, and thus, has a finite number of terms.