Question

The value of the infinite product ${6^{\dfrac{1}{2}}} \times {6^{\dfrac{1}{4}}} \times {6^{\dfrac{1}{8}}} \times {6^{\dfrac{1}{{16}}}} \times ...$ is
A. $6$
B. $36$
C. $216$
D. $\infty $

Answer 1

Hint: We have to find an infinite product in this equation given as ${6^{\dfrac{1}{2}}} \times {6^{\dfrac{1}{4}}} \times {6^{\dfrac{1}{8}}} \times {6^{\dfrac{1}{{16}}}} \times ...$ , we see that all the terms are in exponential form and the base of each term is the same, that is, $6$ . We know that expressions with the same bases can be multiplied using the product rule of exponents. The exponents must be added while keeping the base constant in order to multiply two expressions with the same base, according to this rule. Exponents with the same base must be added when applying this rule. Simplify the given equation using this rule and thus find the infinite product.

Formula used::
Product of the exponents having same base is given as ${a^m} \times {a^n} = {a^{m + n}}$
Sum of an infinte GP is given as $\dfrac{a}{{1 - r}}$

Complete step by step answer:
We have to find the infinite product ${6^{\dfrac{1}{2}}} \times {6^{\dfrac{1}{4}}} \times {6^{\dfrac{1}{8}}} \times {6^{\dfrac{1}{{16}}}} \times ...$
We know that ${a^m} \times {a^n} = {a^{m + n}}$ , so we get:
${6^{\dfrac{1}{2}}} \times {6^{\dfrac{1}{4}}} \times {6^{\dfrac{1}{8}}} \times {6^{\dfrac{1}{{16}}}} \times ... = {6^{(\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{{16}} + ...)}}$ …(1)
To find the value of the infinite product, we have to first find the value of $\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{{16}} + ...$
We see that the first term of the above series is $\dfrac{1}{2}$ and each term is $\dfrac{1}{2}$ times its previous term, so it is an infinite geometric sequence with first term, $a = \dfrac{1}{2}$ and common difference $r = \dfrac{1}{2}$ .
We know that the sum of an infinite geometric sequence is given as $\dfrac{a}{{1 - r}}$ , so the sum of the above geometric sequence is
$
  \dfrac{{\dfrac{1}{2}}}{{1 - \dfrac{1}{2}}} = \dfrac{{\dfrac{1}{2}}}{{\dfrac{{2 - 1}}{2}}} \\
   = \dfrac{{\dfrac{1}{2}}}{{\dfrac{1}{2}}} \\
   = 1 \\
 $
That is, $\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{{16}} + ... = 1$
Now, we put this value in (1):
$
  {6^{\dfrac{1}{2}}} \times {6^{\dfrac{1}{4}}} \times {6^{\dfrac{1}{8}}} \times {6^{\dfrac{1}{{16}}}} \times ... = {6^{(1)}} \\
   \Rightarrow {6^{\dfrac{1}{2}}} \times {6^{\dfrac{1}{4}}} \times {6^{\dfrac{1}{8}}} \times {6^{\dfrac{1}{{16}}}} \times ... = 6 \\
 $
Thus the correct option is option A.

Note:
The sum of a geometric sequence is usually found using the formula ${S_n} = \dfrac{{a({r^n} - 1)}}{{r - 1}}$ , but this formula is for a geometric sequence that has a finite number of terms $n$ . So, we had to apply the formula for the sum of an infinite geometric series in this solution as the number of terms is infinite.