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Question: The value of \[{4^{1/3}}{4^{1/9}}{4^{1/27}}...\infty \]is A. \[2\] B. \[3\] C. \[4\] D. \[9\...

The value of 41/341/941/27...∞{4^{1/3}}{4^{1/9}}{4^{1/27}}...\infty is
A. 22
B. 33
C. 44
D. 99

Explanation

Solution

First, we have to get a series from the power of the given term. For a series having the first termaa, the common ratiorr then their sum isS=a1−rS = \dfrac{a}{{1 - r}}. Using this we can find the value of series and substituting them will give us the required value.
Formula: The sum of an infinite seriesS=a1−rS = \dfrac{a}{{1 - r}} where a$$$$ - the first term of the series, r$$$$ - the common ratio of the series.
Some other formula that we need to know are:
axay=ax+y{a^x}{a^y} = {a^{x + y}}
a×a=a\sqrt {a \times a} = a

Complete step by step answer:
It is given that41/341/941/27...∞{4^{1/3}}{4^{1/9}}{4^{1/27}}...\infty . We aim to find the value of this expression.
Consider the expression41/341/941/27...∞{4^{1/3}}{4^{1/9}}{4^{1/27}}...\infty .
Let us simplify this using the formulaaxay=ax+y{a^x}{a^y} = {a^{x + y}}. Since we have the base as four in all the terms but the power varies.
{4^{1/3}}{4^{1/9}}{4^{1/27}}...\infty $$$$ = {4^{\left( {\dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{{27}} + ...\infty } \right)}}
Now let13+19+127+...∞=x\dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{{27}} + ...\infty = x. Substituting it in the above expression we get.
4(13+19+127+...∞)=4x{4^{\left( {\dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{{27}} + ...\infty } \right)}} = {4^x}
Now let us consider the series13,19,127,...∞\dfrac{1}{3},\dfrac{1}{9},\dfrac{1}{{27}},...\infty . Let’s find the sum of this series.
We know that the sum of the series S=a1−rS = \dfrac{a}{{1 - r}}where a$$$$ - the first term of the series, r$$$$ - the common ratio of the series.
Then for the series13,19,127,...∞\dfrac{1}{3},\dfrac{1}{9},\dfrac{1}{{27}},...\infty , a$$$$ = $$$$\dfrac{1}{3}and r=13r = \dfrac{1}{3}
Thus, the sum of this series S=131−13S = \dfrac{{\dfrac{1}{3}}}{{1 - \dfrac{1}{3}}}
Let us simplify this expression.
S = \dfrac{{\dfrac{1}{3}}}{{\dfrac{{3 - 1}}{3}}}$$$$ = \dfrac{{\dfrac{1}{3}}}{{\dfrac{2}{3}}}$$$$ = \dfrac{1}{2}
Thus, we get that the sum of the series13,19,127,...∞\dfrac{1}{3},\dfrac{1}{9},\dfrac{1}{{27}},...\infty . That is 13+19+127+...∞=12\dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{{27}} + ...\infty = \dfrac{1}{2}
We have that 13+19+127+...∞=x\dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{{27}} + ...\infty = xsox=12x = \dfrac{1}{2}. Substituting this in the expression4(13+19+127+...∞)=4x{4^{\left( {\dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{{27}} + ...\infty } \right)}} = {4^x} we get
{4^{\left( {\dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{{27}} + ...\infty } \right)}} = {4^x}$$$$ = {4^{\dfrac{1}{2}}}
We know that the square root of a number is nothing but the number raised to the power12\dfrac{1}{2}.
Thus, {4^{\left( {\dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{{27}} + ...\infty } \right)}}$$$$ = {4^{\dfrac{1}{2}}} = \sqrt 4
Here four can be written as two into two.
{4^{\left( {\dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{{27}} + ...\infty } \right)}}$$$$ = {4^{\dfrac{1}{2}}} = \sqrt 4 $$$$ = \sqrt {2 \times 2}
Then by using the formula a×a=a\sqrt {a \times a} = awe get
{4^{\left( {\dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{{27}} + ...\infty } \right)}}$$$$ = 2

So, the correct answer is “Option A”.

Note: The sum of the infinite series can be found by using the standard formula. Here we got the power as a sum of series thus, we calculated its value by using the standard formula. The series that we got from the given expression is infinite thus, we used this formula.