Question

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} $

Answer 1

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.