Question

Prove that \[{9^{\dfrac{1}{3}}} \times {9^{\dfrac{1}{9}}} \times {9^{\dfrac{1}{{27}}}} \times ...\infty = 3\].

Answer 1

Hint: The given number is given in the form of exponential, here the base value is the same. By using the law of indices we rewrite the given expression and the series we will obtain. So by using the definition of series we are going to simplify further and hence we prove the given question.

Complete step by step solution:
The exponential number is defined as the number of times the number is multiplied by itself. It is represented as \[{a^n}\], where a is the numeral and n represents the number of times the number is multiplied.
Now consider the given question
\[ {9^{\dfrac{1}{3}}} \times {9^{\dfrac{1}{9}}} \times {9^{\dfrac{1}{{27}}}} \times ...\infty = 3\]
On considering the LHS we have
\[ \Rightarrow {9^{\dfrac{1}{3}}} \times {9^{\dfrac{1}{9}}} \times {9^{\dfrac{1}{{27}}}} \times ...\infty \]
By using the law of indices \[{a^m} \times {a^n} = {a^{m + n}}\], the above inequality is written as
\[ \Rightarrow {9^{\left( {\dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{{27}} + ...\infty } \right)}}\]--------------(1)
The power or exponent term is in the form of geometric series, Since it is in the form of summation.
We have formula for the sum of the G.P we have two kind of depending on the common ratio and that is defined as
If the common ratio greater than 1 we have \[{S_n} = \dfrac{{a({r^n} - 1)}}{{r - 1}}\]and if the common ratio is less than 1 we have \[{S_n} = \dfrac{{a(1 - {r^n})}}{{1 - r}}\]
When we have to find the sum for the infinite series we use the formula \[{S_n} = \dfrac{a}{{1 - r}}\]
Now consider the series \[\dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{{27}} + ...\infty \]
So we have the first term \[a = \dfrac{1}{3}\] and \[r = \dfrac{1}{3}\]
\[ \Rightarrow {S_n} = \dfrac{a}{{1 - r}}\]
On substituting the values we have
\[ \Rightarrow {S_n} = \dfrac{{\dfrac{1}{3}}}{{1 - \dfrac{1}{3}}}\]
On simplifying we have
\[ \Rightarrow {S_n} = \dfrac{{\dfrac{1}{3}}}{{\dfrac{2}{3}}}\]
On further simplifying we have
\[ \Rightarrow {S_n} = \dfrac{1}{2}\]
This is summation is
\[ \Rightarrow \dfrac{1}{3} + \dfrac{1}{9} + \dfrac{1}{{27}} + ...\infty = \dfrac{1}{2}\]---------(2)
On substituting (2) in (1) we have
\[ \Rightarrow {9^{\dfrac{1}{2}}}\]
This can be written as
\[ \Rightarrow \sqrt 9 \]
\[ \Rightarrow 3\]
Hence \[{9^{\dfrac{1}{3}}} \times {9^{\dfrac{1}{9}}} \times {9^{\dfrac{1}{{27}}}} \times ...\infty = 3\]
Hence proved.

Note:
We must know about the geometric progression arrangement and it is based on the first term and common ratio. The common ratio of the geometric progression is defined as \[\dfrac{{{a_2}}}{{{a_1}}}\] where \[{a_2}\] represents the second term and \[{a_1}\] represents the first term. The sum of n terms is defined on the basis of common ratio.