Question

The value of the infinite product ${6^{\dfrac{1}{2}}} \times {6^{\dfrac{1}{2}}} \times {6^{\dfrac{3}{8}}} \times {6^{\dfrac{1}{4}}} \ldots \ldots$is.
A. 6
B. 36
C. 216
D. $\infty$

Answer 1

We will first consider the given expression and transform the powers in a way such that it forms the pattern. After making the pattern add the powers which have the same base and determine whether the pattern is arithmetic series or geometric series. As the series is an arithmetic series find the common difference using $d = {a_2} - {a_1}$ and write the equation and after this multiply both sides by common difference and subtract it from the original equation which gives the geometric series. Now, determine the value of the common ratio using $r = \dfrac{{{a_2}}}{{{a_1}}}$ and find the sum of infinite series using $\dfrac{1}{{1 - r}}$ after this substitute the value obtained of sum and find the product of the series.

Complete step by step answer:

Consider the given expression that is ${6^{\dfrac{1}{2}}} \times {6^{\dfrac{1}{2}}} \times {6^{\dfrac{3}{8}}} \times {6^{\dfrac{1}{4}}} \ldots \ldots$
This equation can be rewritten as ${6^{\dfrac{1}{2}}} \times {6^{\dfrac{2}{4}}} \times {6^{\dfrac{3}{8}}} \times {6^{\dfrac{4}{{16}}}} \ldots \ldots$
Now, we will add the powers as all the factors have the same base.
Thus, we get,
$\Rightarrow {6^{\left( {\dfrac{1}{2} + \dfrac{2}{4} + \dfrac{3}{8} + \dfrac{4}{{16}} + \ldots \ldots } \right)}}$
Now, we will let $s = \dfrac{1}{2} + \dfrac{2}{4} + \dfrac{3}{8} + \dfrac{4}{{16}} + \ldots$,
Thus, the infinite product will be of the form ${6^s}$.
Which is an arithmetic-geometric progression, where the denominator raises a G.P and the numerator raise as an A.P.
Next we will write the denominator in the form of powers having base as 2.
thus, we get,
$s = \dfrac{1}{2} + \dfrac{2}{{{2^2}}} + \dfrac{3}{{{2^3}}} + \dfrac{4}{{{2^4}}} + \ldots \;\;\;\;\;\; \to \left( 1 \right)$
next, we will find the common difference which is represented by $d$ and the formula is $d = {a_2} - {a_1}$:
Hence, we have,
$\Rightarrow \dfrac{1}{{{2^2}}} - \dfrac{1}{2} = \dfrac{1}{2}$
Further, Multiply both sides of the equation $s = \dfrac{1}{2} + \dfrac{2}{4} + \dfrac{3}{8} + \dfrac{4}{{16}} + \ldots$ by $d$, that is $\dfrac{1}{2}$.
thus, we get,
$\Rightarrow \dfrac{s}{2} = \dfrac{1}{{{2^2}}} + \dfrac{2}{{{2^3}}} + \dfrac{3}{{{2^4}}} + \ldots \;\;\;\;\;\; \to \left( 2 \right)$
Next, we will Subtract equation (1) from equation (2),

$$\Rightarrow s - \dfrac{s}{2} = \left( {\dfrac{1}{2} + \dfrac{2}{{{2^2}}} + \dfrac{3}{{{2^3}}} + \dfrac{4}{{{2^4}}} + \ldots } \right) - \left( {\dfrac{1}{{{2^2}}} + \dfrac{2}{{{2^3}}} + \dfrac{3}{{{2^4}}} + \ldots } \right) \\\ \Rightarrow \dfrac{s}{2} = \dfrac{1}{2} + \dfrac{1}{{{2^2}}} + \dfrac{1}{{{2^3}}} + \ldots \\\$$

Which becomes a geometric series of infinite terms.
Now, we will find the common ratio of the series using $r = \dfrac{{{a_2}}}{{{a_1}}}$ which is represented by $r$,
The common ratio of the series is $r = \dfrac{1}{2}$
Further, we will apply the formula of the sum of an infinite geometric series which is given by $s = \dfrac{1}{{1 - r}}$ Thus, we have,
$\Rightarrow \dfrac{s}{2} = \dfrac{{\dfrac{1}{2}}}{{1 - \dfrac{1}{2}}} = 1$
Further, we will solve for $s$.
$\Rightarrow s = 2$
Next, we will substitute the value of $s$ in ${6^s}$.
Thus, we get,
$\Rightarrow {6^s} = {6^2} = 36$
So, the product of ${6^{\dfrac{1}{2}}} \times {6^{\dfrac{1}{2}}} \times {6^{\dfrac{3}{8}}} \times {6^{\dfrac{1}{4}}} \ldots \ldots$ is 36.
Hence, option (B) is correct.

Note: To convert the arithmetic series into geometric series, multiply the common ratio with arithmetic series and subtract it form the original series, do not add the equations as it will make the series large and complicated. Use the formula for common ration and common difference to evaluate the values. Apply the formula for the sum of infinite series of geometric series.