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# Find the value of $\sum\limits_{k=1}^{20}{\dfrac{1}{{{2}^{k}}}}$ is equal to?  \begin{align} & a)2-\dfrac{11}{{{2}^{19}}} \\ & b)1-\dfrac{1}{{{2}^{20}}} \\ & c)2-\dfrac{1}{{{2}^{20}}} \\ & d)1-\dfrac{11}{{{2}^{20}}} \\ \end{align}

Last updated date: 20th Jun 2024
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Hint: Now we know that $\sum\limits_{k=1}^{20}{\dfrac{1}{{{2}^{k}}}}=\dfrac{1}{2}+\dfrac{1}{{{2}^{2}}}+\dfrac{1}{{{2}^{3}}}+.....+\dfrac{1}{{{2}^{20}}}$ Now this is a GP with first term as $\dfrac{1}{2}$ and common difference as $\dfrac{1}{2}$ and we know that the sum of n terms of GP with first term a and common difference r is given by $\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}$ . hence using this formula we can find the sum of GP.

Now consider $\sum\limits_{k=1}^{20}{\dfrac{1}{{{2}^{k}}}}$ we know that $\sum\limits_{k=1}^{20}{\dfrac{1}{{{2}^{k}}}}=\dfrac{1}{2}+\dfrac{1}{{{2}^{2}}}+\dfrac{1}{{{2}^{3}}}+...+\dfrac{1}{{{2}^{20}}}$
Now consider the sequence $\dfrac{1}{2},\dfrac{1}{{{2}^{2}}},\dfrac{1}{{{2}^{3}}},\dfrac{1}{{{2}^{4}}}...\dfrac{1}{{{2}^{20}}}$
$\Rightarrow$ The first term of the sequence is $\dfrac{1}{2}$ .
$\Rightarrow$ Now if we see each term is multiplied by $\dfrac{1}{2}$ to obtain next term.
$\Rightarrow$ Hence this Sequence is in the form of $a,ar,a{{r}^{2}},a{{r}^{3}}...$
But such a sequence is called GP.
Hence we have the given sequence is a GP.
$\Rightarrow$ Now we know that for any GP with its first term a and common ratio r sum of n terms is given by formula $\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}$ if r is not equal to 1.
Now again consider our GP $\dfrac{1}{2},\dfrac{1}{{{2}^{2}}},\dfrac{1}{{{2}^{3}}},\dfrac{1}{{{2}^{4}}}...\dfrac{1}{{{2}^{20}}}$
In this series we know that the in our GP first term is $\dfrac{1}{2}$ and the common difference is also $\dfrac{1}{2}$
Hence substituting n = 20, $a=\dfrac{1}{2}$ and $r=\dfrac{1}{2}$ in the formula $\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}$ we get.
${{S}_{20}}=\dfrac{\left( \dfrac{1}{2} \right)\left( 1-{{\left( \dfrac{1}{2} \right)}^{20}} \right)}{1-\dfrac{1}{2}}$
Now let us simplify the above expression.
\begin{align} & {{S}_{20}}=\dfrac{\left( \dfrac{1}{2} \right)\left( 1-{{\left( \dfrac{1}{2} \right)}^{20}} \right)}{\dfrac{2-1}{2}} \\ & =\dfrac{2\times \left( \dfrac{1}{2} \right)\left( 1-{{\left( \dfrac{1}{2} \right)}^{20}} \right)}{1} \\ & =1-\dfrac{1}{{{2}^{20}}} \\ \end{align}
$\Rightarrow$ Hence we get sum of 20 terms of the GP is $1-\dfrac{1}{{{2}^{20}}}$
$\Rightarrow$ Hence we get the value of $\sum\limits_{k=1}^{20}{\dfrac{1}{{{2}^{k}}}}$ is equal to $1-\dfrac{1}{{{2}^{20}}}$ .
$\Rightarrow$ Option b is the correct option.

Note:
Now Note that the sum of GP is given by $\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}$ Now this formula is only when r is not equal to 1. If we have r = 1 then the sequence is constant sequence and sum is given by na.