Question

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

Answer 1

Hint: First, we have to get a series from the power of the given term. For a series having the first term\[a\], the common ratio\[r\] then their sum is\[S = \dfrac{a}{{1 - r}}\]. Using this we can find the value of series and substituting them will give us the required value.
Formula: The sum of an infinite series\[S = \dfrac{a}{{1 - r}}\] where \[a\]\[ - \]the first term of the series, \[r\]\[ - \]the common ratio of the series.
Some other formula that we need to know are:
\[{a^x}{a^y} = {a^{x + y}}\]
\[\sqrt {a \times a} = a\]

Complete step by step answer:
It is given that\[{4^{1/3}}{4^{1/9}}{4^{1/27}}...\infty \]. We aim to find the value of this expression.
Consider the expression\[{4^{1/3}}{4^{1/9}}{4^{1/27}}...\infty \].
Let us simplify this using the formula\[{a^x}{a^y} = {a^{x + y}}\]. 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 let\[\dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{{27}} + ...\infty = x\]. Substituting it in the above expression we get.
\[{4^{\left( {\dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{{27}} + ...\infty } \right)}} = {4^x}\]
Now let us consider the series\[\dfrac{1}{3},\dfrac{1}{9},\dfrac{1}{{27}},...\infty \]. Let’s find the sum of this series.
We know that the sum of the series \[S = \dfrac{a}{{1 - r}}\]where \[a\]\[ - \]the first term of the series, \[r\]\[ - \]the common ratio of the series.
Then for the series\[\dfrac{1}{3},\dfrac{1}{9},\dfrac{1}{{27}},...\infty \], \[a\]\[ = \]\[\dfrac{1}{3}\]and \[r = \dfrac{1}{3}\]
Thus, the sum of this series \[S = \dfrac{{\dfrac{1}{3}}}{{1 - \dfrac{1}{3}}}\]
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 series\[\dfrac{1}{3},\dfrac{1}{9},\dfrac{1}{{27}},...\infty \]. That is \[\dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{{27}} + ...\infty = \dfrac{1}{2}\]
We have that \[\dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{{27}} + ...\infty = x\]so\[x = \dfrac{1}{2}\]. Substituting this in the expression\[{4^{\left( {\dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{{27}} + ...\infty } \right)}} = {4^x}\] 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 power\[\dfrac{1}{2}\].
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 \[\sqrt {a \times a} = a\]we 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.