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# The product $\left( {32} \right){\left( {32} \right)^{\dfrac{1}{6}}}{\left( {32} \right)^{\dfrac{1}{{36}}}}.......\infty$ isA. $16$ B. $32$ C. $64$ D. $0$

Last updated date: 13th Jun 2024
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Hint: Here, the given term is in geometric progression as the terms are increasing in fixed ratio. So, we will use the concept of Geometric Progression to solve the question. A geometric progression is a sequence or series of numbers where each term after the first is found out by multiplying the previous one by a fixed number called the common ratio.

Formula used:
We will use the following formulas:
1. Exponential Formula: ${a^m} \cdot {a^n} \cdot {a^o} \cdot {a^p}..... = {a^{m + n + o + p + .........}}$
2. Exponential Formula: ${\left( {{a^m}} \right)^n} = {a^{mn}}$
3. Geometric Progression is given by ${S_n} = \dfrac{{a(1 - {r^n})}}{{(1 - r)}}$ , where $a$ is the first term and $r$ is the common ratio.

Complete Step by Step Solution:
We are given with a geometric Series $\left( {32} \right){\left( {32} \right)^{\dfrac{1}{6}}}{\left( {32} \right)^{\dfrac{1}{{36}}}}.......\infty$.
The given series is of the form $a \cdot ar \cdot a{r^2} \cdot ....... \cdot a{r^n}$
Thus, the first term of the Geometric Series $a = 32$ and $r = {1^{\dfrac{1}{6}}}$.
By using the formula ${a^m} \cdot {a^n} \cdot {a^o} \cdot {a^p}..... = {a^{m + n + o + p + .........}}$, we can rewrite the given equation as:
$\Rightarrow \left( {32} \right){\left( {32} \right)^{\dfrac{1}{6}}}{\left( {32} \right)^{\dfrac{1}{{36}}}}.......\infty = {32^{\left( {1 + \dfrac{1}{6} + \dfrac{1}{{36}} + .......} \right)}}$
Now, by applying the formula of Geometric Progression ${S_n} = \dfrac{{a(1 - {r^n})}}{{(1 - r)}}$ to the power, we get
$\Rightarrow \left( {32} \right){\left( {32} \right)^{\dfrac{1}{6}}}{\left( {32} \right)^{\dfrac{1}{{36}}}}.......\infty = {32^{\left( {\dfrac{{1(1 - 0)}}{{1 - \dfrac{1}{6}}}} \right)}}$
By cross multiplying, we get
$\Rightarrow \left( {32} \right){\left( {32} \right)^{\dfrac{1}{6}}}{\left( {32} \right)^{\dfrac{1}{{36}}}}.......\infty = {32^{\left( {\dfrac{1}{{\dfrac{{6 - 1}}{6}}}} \right)}}$
Simplifying the expression, we get
$\Rightarrow \left( {32} \right){\left( {32} \right)^{\dfrac{1}{6}}}{\left( {32} \right)^{\dfrac{1}{{36}}}}.......\infty = {32^{\left( {\dfrac{1}{{\dfrac{5}{6}}}} \right)}}$
$\Rightarrow \left( {32} \right){\left( {32} \right)^{\dfrac{1}{6}}}{\left( {32} \right)^{\dfrac{1}{{36}}}}.......\infty = {32^{\left( {\dfrac{6}{5}} \right)}}$
Rewriting $32$ in terms of the power of $2$, we get
$\Rightarrow \left( {32} \right){\left( {32} \right)^{\dfrac{1}{6}}}{\left( {32} \right)^{\dfrac{1}{{36}}}}.......\infty = {2^{^5}}^{\left( {\dfrac{6}{5}} \right)}$
Now, by using the formula ${\left( {{a^m}} \right)^n} = {a^{mn}}$, we have
$\Rightarrow \left( {32} \right){\left( {32} \right)^{\dfrac{1}{6}}}{\left( {32} \right)^{\dfrac{1}{{36}}}}.......\infty = {2^6}$
$\Rightarrow \left( {32} \right){\left( {32} \right)^{\dfrac{1}{6}}}{\left( {32} \right)^{\dfrac{1}{{36}}}}.......\infty = 64$

Therefore, the product $\left( {32} \right){\left( {32} \right)^{\dfrac{1}{6}}}{\left( {32} \right)^{\dfrac{1}{{36}}}}.......\infty$ is $64$.

Note:
Here, we need to remember the basics of the Geometric Series and Geometric Sequence. The properties of G.P. are:
1. If every term of G.P. is multiplied or divided by a non-zero number, then the resulting terms are also in G.P.
2. If the common ratio is negative, then the result will alternate between positive and negative.
3. If the common ratio is greater than 1 then there will be an exponential growth towards infinity (positive).
4. If the common ratio is less than $-1$ then there will be an exponential growth towards infinity (positive and negative).