Question

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) $1 + \dfrac{1}{{{2^n}}}$

Answer 1

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

\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.