Question

The value of ${{2}^{\dfrac{1}{4}}}{{.4}^{\dfrac{1}{8}}}{{.8}^{\dfrac{1}{16}}}......$ infinity is
1. $1$
2. $2$
3. $\dfrac{3}{2}$
4. $4$

Answer 1

Firstly convert the given series into the same base so that you can apply the product of powers rule. After that try to convert the expression into some kind of series either AP or GP then use the summation formula for that particular series. By using these steps you can solve the question.

Complete step by step answer:
To solve this question firstly let us understand what the exponent is and what are the rules to solve the exponent.
Exponents also called powers are the way of performing the repeated multiplications. For example if we want to multiply $2$ , $5$ times. Then there are the two methods of representing,
The first one is $2\times 2\times 2\times 2\times 2$ which is called the normal way of multiplication and the second one is ${{2}^{5}}$ which is known as the exponent way of solving the multiplication.
In ${{2}^{5}}$, $2$represents the base i.e. the number that gets multiplied and $5$represent the exponent i.e. the number of times the base is multiplied by itself.
Rules to solve the exponents are:
Product of powers: When the bases are in product then add the powers together. For example: ${{2}^{2}}\times {{2}^{4}}\times {{2}^{6}}\times {{2}^{3}}={{2}^{2+4+6+3}}$
Division of powers: When the bases are in division then subtract the powers. For example: $\dfrac{{{2}^{3}}}{{{2}^{2}}}={{2}^{3-2}}$
Power of powers: When one exponent is raised with another exponent then multiply powers together. For example: ${{4}^{{{2}^{3}}}}={{4}^{2\times 3}}$
Zero power: If any base is raised with the power of zero then it becomes $1$. For example: ${{6}^{0}}=1$
After applying these exponent rules, check which series is hidden in the given question. The series can either be Arithmetic Progression or Geometric progression. The series is an AP series if the common difference is the same for the consecutive numbers. And the series is a GP series if the common ratio between the consecutive numbers is the same.
Now let us solve the question,
We have to find the value of ${{2}^{\dfrac{1}{4}}}{{.4}^{\dfrac{1}{8}}}{{.8}^{\dfrac{1}{16}}}......$ infinity. Our first task is to make the bases same
${{2}^{{{1}^{\dfrac{1}{4}}}}}\times {{2}^{{{2}^{\dfrac{1}{8}}}}}\times {{2}^{{{3}^{\dfrac{1}{16}}}}}......\infty$
Now using power of powers rule, we get
$\Rightarrow {{2}^{\dfrac{1}{4}}}\times {{2}^{\dfrac{2}{8}}}\times {{2}^{\dfrac{3}{16}}}......\infty$
Using the rule product of powers, we get
$\Rightarrow {{2}^{\dfrac{1}{4}+\dfrac{2}{8}+\dfrac{3}{16}......\infty }}$
Let us solve this summation series individually. Assume the series as $S$
$\Rightarrow S=\dfrac{1}{4}+\dfrac{2}{8}+\dfrac{3}{16}......\infty$ $........(1)$
Now divide the equation $(1)$ by $2$ and we get
\Rightarrow \dfrac{S}{2}=\dfrac{1}{8}+\dfrac{2}{16}+\dfrac{3}{32}......\infty $$$$.........(2)
Subtract equation $(1)$ and $(2)$,
$\Rightarrow \dfrac{S}{2}=\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}......\infty$
Now if we observe that the series formed is the GP series because the common ratio $r$ is same i.e. $\dfrac{1}{2}$.
The summation formula for the infinite GP series is $=\dfrac{a}{1-r}$
$\Rightarrow \dfrac{S}{2}=\dfrac{\dfrac{1}{4}}{1-\dfrac{1}{2}}$

& \Rightarrow \dfrac{S}{2}=\dfrac{1}{2} \\\ & \Rightarrow S=1 \\\ \end{aligned}$$ Now let us put the value $$S$$at which we assume, we get $$\begin{aligned} & \Rightarrow {{2}^{S}} \\\ & \Rightarrow {{2}^{1}} \\\ & \Rightarrow 2 \\\ \end{aligned}$$ **So, the correct answer is “Option 2”.** **Note:** If we are given the series of $$n$$ natural numbers then their summation will be $$\dfrac{n(n+1)}{2}$$. And if we have the series of $$n$$ even numbers then their summation will be $$n(n+1)$$, where $$n$$ is the number of digits in the series. And for the $$n$$ odd numbers the summation will be $${{n}^{2}}$$.