
Check if the value of $\left( {\dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{{16}} + \ldots \ldots + \infty } \right)$ is $\dfrac{1}{2}$
Answer
609k+ views
Hint: Try to take out common terms and make the series simpler to solve.
Given series: $\left( {\dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{{16}} + \ldots \ldots + \infty } \right)$
$ \Rightarrow \dfrac{1}{4}\left[ {1 + \dfrac{1}{2} + \dfrac{1}{4} + \ldots \ldots + \infty } \right]{\text{ }} \ldots \left( 1 \right)$
Here, we assume the above equation as the sum of terms in a Geometric Progression with $a = 1,{\text{ }}r = \dfrac{{\dfrac{1}{2}}}{1} = 1$ .
Now, applying the sum of a Geometric Progression:
$
\Rightarrow \left[ {1 + \dfrac{1}{2} + \dfrac{1}{4} + \ldots \ldots + \infty } \right] = \dfrac{a}{{1 - r}} \\
\Rightarrow \left[ {1 + \dfrac{1}{2} + \dfrac{1}{4} + \ldots \ldots + \infty } \right] = \dfrac{1}{{\left( {1 - \dfrac{1}{2}} \right)}} \\
\Rightarrow \left[ {1 + \dfrac{1}{2} + \dfrac{1}{4} + \ldots \ldots + \infty } \right] = \dfrac{1}{{\left( {\dfrac{1}{2}} \right)}} \\
\Rightarrow \left[ {1 + \dfrac{1}{2} + \dfrac{1}{4} + \ldots \ldots + \infty } \right] = 2 \\
$
Putting this value in equation (1), we get
$
\Rightarrow \dfrac{1}{4}\left( 2 \right) \\
\Rightarrow \dfrac{1}{2} \\
$
Note: Whenever we are supposed to find the sum of a series, always try to make the series in the form of Arithmetic Progression or Geometric Progression or Harmonic Progression and then simply use the sum of series formula which is already defined for these progressions.
Given series: $\left( {\dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{{16}} + \ldots \ldots + \infty } \right)$
$ \Rightarrow \dfrac{1}{4}\left[ {1 + \dfrac{1}{2} + \dfrac{1}{4} + \ldots \ldots + \infty } \right]{\text{ }} \ldots \left( 1 \right)$
Here, we assume the above equation as the sum of terms in a Geometric Progression with $a = 1,{\text{ }}r = \dfrac{{\dfrac{1}{2}}}{1} = 1$ .
Now, applying the sum of a Geometric Progression:
$
\Rightarrow \left[ {1 + \dfrac{1}{2} + \dfrac{1}{4} + \ldots \ldots + \infty } \right] = \dfrac{a}{{1 - r}} \\
\Rightarrow \left[ {1 + \dfrac{1}{2} + \dfrac{1}{4} + \ldots \ldots + \infty } \right] = \dfrac{1}{{\left( {1 - \dfrac{1}{2}} \right)}} \\
\Rightarrow \left[ {1 + \dfrac{1}{2} + \dfrac{1}{4} + \ldots \ldots + \infty } \right] = \dfrac{1}{{\left( {\dfrac{1}{2}} \right)}} \\
\Rightarrow \left[ {1 + \dfrac{1}{2} + \dfrac{1}{4} + \ldots \ldots + \infty } \right] = 2 \\
$
Putting this value in equation (1), we get
$
\Rightarrow \dfrac{1}{4}\left( 2 \right) \\
\Rightarrow \dfrac{1}{2} \\
$
Note: Whenever we are supposed to find the sum of a series, always try to make the series in the form of Arithmetic Progression or Geometric Progression or Harmonic Progression and then simply use the sum of series formula which is already defined for these progressions.
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