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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.

**Complete step by step answer:**

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.

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