Question

Find the value of ${{2}^{\dfrac{1}{4}}}{{.4}^{\dfrac{1}{8}}}{{.8}^{\dfrac{1}{16}}}{{.16}^{\dfrac{1}{32}}}......$?
(a) 1
(b) 2
(c) $\dfrac{3}{2}$
(d) $\dfrac{5}{2}$

Answer 1

We start solving the problem by writing 4, 8, 16 in terms of exponents of 2. We then use the law of exponent ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$ for the exponents, we just obtained. We then use the law of exponent ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$. We then find the sum of the terms present in the power of 2 by making necessary calculations. After finding the sum, we use it and make the required calculations to get the required answer.

Complete step by step answer:
According to the problem, we need to find the value of ${{2}^{\dfrac{1}{4}}}{{.4}^{\dfrac{1}{8}}}{{.8}^{\dfrac{1}{16}}}{{.16}^{\dfrac{1}{32}}}......$.
Let us assume the value of ${{2}^{\dfrac{1}{4}}}{{.4}^{\dfrac{1}{8}}}{{.8}^{\dfrac{1}{16}}}{{.16}^{\dfrac{1}{32}}}......$ be ‘x’.
So, we have $x={{2}^{\dfrac{1}{4}}}{{.4}^{\dfrac{1}{8}}}{{.8}^{\dfrac{1}{16}}}{{.16}^{\dfrac{1}{32}}}......$.
$\Rightarrow x={{2}^{\dfrac{1}{4}}}.{{\left( {{2}^{2}} \right)}^{\dfrac{1}{8}}}.{{\left( {{2}^{3}} \right)}^{\dfrac{1}{16}}}.{{\left( {{2}^{4}} \right)}^{\dfrac{1}{32}}}......$---(1).
From laws of exponents, we know that ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$. Let us use this in equation (1).
$\Rightarrow x={{2}^{\dfrac{1}{4}}}{{.2}^{\dfrac{2}{8}}}{{.2}^{\dfrac{3}{16}}}{{.2}^{\dfrac{4}{32}}}......$ ---(2).
From laws of exponents, we know that ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$. Let us use this in equation (2).
$\Rightarrow x={{2}^{\dfrac{1}{4}+\dfrac{2}{8}+\dfrac{3}{16}+\dfrac{4}{32}+.....}}$ ---(3).
Let us first find the sum of $\dfrac{1}{4}+\dfrac{2}{8}+\dfrac{3}{16}+\dfrac{4}{32}+.....$. Let us assume the sum as S.
So, we have $x={{2}^{S}}$ ---(4)
So, we have $S=\dfrac{1}{4}+\dfrac{2}{8}+\dfrac{3}{16}+\dfrac{4}{32}+.....$ ---(5). Let us divide this S with 2.
We get $\dfrac{S}{2}=\dfrac{1}{8}+\dfrac{2}{16}+\dfrac{3}{32}+\dfrac{4}{64}+.....$ ---(6).
Let us subtract equation (6) from equation (5).
So, $S-\dfrac{S}{2}=\left( \dfrac{1}{4}+\dfrac{2}{8}+\dfrac{3}{16}+\dfrac{4}{32}+..... \right)-\left( \dfrac{1}{8}+\dfrac{2}{16}+\dfrac{3}{32}+\dfrac{4}{64}+..... \right)$.
$\Rightarrow \dfrac{S}{2}=\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+.....$.
$\Rightarrow \dfrac{S}{2}=\dfrac{1}{4}\left( 1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+..... \right)$.
Here we can see that $1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+.....$ resembles infinite geometric series with first term ‘1’ and common ration $\dfrac{1}{2}$.
We know that sum of infinite geometric series is $\dfrac{a}{1-r}\left( if\left| r \right|<1 \right)$, where ‘a’ is first term and ‘r’ is common ratio.
$\Rightarrow \dfrac{S}{2}=\dfrac{1}{4}\left( \dfrac{1}{1-\dfrac{1}{2}} \right)$.
$\Rightarrow \dfrac{S}{2}=\dfrac{1}{4}\left( \dfrac{1}{\dfrac{1}{2}} \right)$.
$\Rightarrow \dfrac{S}{2}=\dfrac{1}{4}\left( 2 \right)$.
$\Rightarrow \dfrac{S}{2}=\dfrac{1}{2}$.
$\Rightarrow S=1$. Let us substitute in equation (4).
So, we get $x={{2}^{1}}=2$.
We have found the value of ${{2}^{\dfrac{1}{4}}}{{.4}^{\dfrac{1}{8}}}{{.8}^{\dfrac{1}{16}}}{{.16}^{\dfrac{1}{32}}}......$ as 2.

So, the correct answer is “Option b”.

Note: We can see that the sum $S=\dfrac{1}{4}+\dfrac{2}{8}+\dfrac{3}{16}+\dfrac{4}{32}+.....$ resembles the infinite sum of arithmetic-geometric series. We can find this sum by using the formula of sum of infinite terms of arithmetic-geometric series $\dfrac{a}{1-r}+\dfrac{dr}{{{\left( 1-r \right)}^{2}}},for\left| r \right|<1$. We should not confuse or make calculation mistakes while solving this problem. Whenever we get this type of problem, we need to make use of laws of exponents which makes our calculations easier.