Question

The value of the series ${{2}^{\dfrac{1}{4}}}{{.4}^{\dfrac{1}{8}}}{{.8}^{\dfrac{1}{16}}}{{.16}^{\dfrac{1}{32}}}........$ is equal to
1). $1$
2). $2$
3). $\dfrac{3}{2}$
4). $\dfrac{5}{2}$

Answer 1

To solve this problem, first we need to convert all the terms in the same base and after that exponents form the geometric progression, and now we find the sum of the geometric progression and then substitute its value and you will get your required answer.

Complete step-by-step solution:
A series can simply be defined as the sum of the various numbers, or elements of a sequence. The series can be finite or infinite depending on the sequence whether it is finite or infinite.
Sequence can be defined as the set of the elements that follow a certain pattern whereas series can be defined as the sum of elements of the given sequence. The finite series are series where the numbers are ending and infinite series are the series where the numbers are never ending.
Types of series are as follows:
$\bullet$ Geometric Series
$\bullet$ Harmonic Series
$\bullet$ Power Series
$\bullet$ Alternating Series
$\bullet$ Exponent Series
A geometric series can be defined as a series with a constant ratio between successive terms.
A harmonic series can be defined as the series that contains the sum of terms that are the reciprocal of the arithmetic series terms.
Power series can be defined as the series that can be thought of as a polynomial with an infinite number of terms.
As we are given in the question:
${{2}^{\dfrac{1}{4}}}{{.4}^{\dfrac{1}{8}}}{{.8}^{\dfrac{1}{16}}}{{.16}^{\dfrac{1}{32}}}........$
$\Rightarrow {{2}^{\dfrac{1}{4}}}.{{({{2}^{2}})}^{\dfrac{1}{8}}}.{{({{2}^{3}})}^{\dfrac{1}{16}}}.{{({{2}^{4}})}^{\dfrac{1}{32}}}........$
$\Rightarrow {{2}^{\dfrac{1}{4}}}.{{(2)}^{\dfrac{2}{8}}}.{{(2)}^{\dfrac{3}{16}}}.{{(2)}^{\dfrac{4}{32}}}........$
After converting all bases into bases of $2$ and then we will use the property that when the base are same and all of them are in multiplication then we will add the exponents together and we will get as:
$\Rightarrow{2}^{{\dfrac{1}{4}+\dfrac{2}{8}+\dfrac{3}{16}+\dfrac{4}{32}}........}$ $.......(1)$
Now, we need to find the sum of this geometric progression
$\Rightarrow S=\dfrac{1}{4}+\dfrac{2}{8}+\dfrac{3}{16}+..........$
Dividing this by $2$
$\Rightarrow \dfrac{S}{2}=\dfrac{1}{8}+\dfrac{2}{16}+\dfrac{3}{32}+..........$
Now subtracting $S$ and $\dfrac{S}{2}$ and we will get:
$\Rightarrow \dfrac{S}{2}=\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}..........$
As this represents the infinite geometric progression in which first term $a=\dfrac{1}{4}$ and constant ratio between two terms $r=\dfrac{1}{2}$
So, $S=\dfrac{a}{1-r}$
$\Rightarrow \dfrac{S}{2}=\dfrac{\dfrac{1}{4}}{\left( 1-\dfrac{1}{2} \right)}$
$\Rightarrow \dfrac{S}{2}=\dfrac{\dfrac{1}{4}}{\left( \dfrac{2-1}{2} \right)}$
$\Rightarrow \dfrac{S}{2}=\dfrac{1}{2}$
$\Rightarrow S=\dfrac{2}{2}$
$\Rightarrow S=1$
Now, we will substitute this value of $S$ in equation $(1)$
$\Rightarrow {{2}^{1}}$
$\Rightarrow 2$
Hence, the correct option is $2$

Note: There is a series called the Fibonacci series in which when we add the last two numbers, then the sum of the last two numbers is treated as the next number in the series ( i.e., each number is obtained by adding the two preceding numbers. Using the Pascal’s triangle, Fibonacci numbers can be obtained.