Q71 / 180 · COMEDK 2024 (Morning Shift) · Mathematics
$(32) \times(32)^{\dfrac{1}{6}} \times(32)^{\dfrac{1}{36}} \times-----\infty \text { is equal to }$
- Correct answerA64
- B16
- C32
- D0
Solution
Let the given expression be $S = (32) \times (32)^{\dfrac{1}{6}} \times (32)^{\dfrac{1}{36}} \times \dots \infty$.
The expression can be written as $S = (32)^{1 + \dfrac{1}{6} + \dfrac{1}{36} + \dots \infty}$.
The exponent is an infinite geometric series with first term $a = 1$ and common ratio $r = \dfrac{1}{6}$.
The sum of an infinite geometric series is given by $\dfrac{a}{1 - r}$, provided $|r| < 1$.
Substituting the values, the sum of the exponent is $\dfrac{1}{1 - \dfrac{1}{6}} = \dfrac{1}{\dfrac{5}{6}} = \dfrac{6}{5}$.
Therefore, $S = (32)^{\dfrac{6}{5}}$.
Since $32 = 2^5$, we have $S = (2^5)^{\dfrac{6}{5}} = 2^{5 \times \dfrac{6}{5}} = 2^6$.
$2^6 = 64$.
Answer: 64
The expression can be written as $S = (32)^{1 + \dfrac{1}{6} + \dfrac{1}{36} + \dots \infty}$.
The exponent is an infinite geometric series with first term $a = 1$ and common ratio $r = \dfrac{1}{6}$.
The sum of an infinite geometric series is given by $\dfrac{a}{1 - r}$, provided $|r| < 1$.
Substituting the values, the sum of the exponent is $\dfrac{1}{1 - \dfrac{1}{6}} = \dfrac{1}{\dfrac{5}{6}} = \dfrac{6}{5}$.
Therefore, $S = (32)^{\dfrac{6}{5}}$.
Since $32 = 2^5$, we have $S = (2^5)^{\dfrac{6}{5}} = 2^{5 \times \dfrac{6}{5}} = 2^6$.
$2^6 = 64$.
Answer: 64
Hence, the correct answer is option: 'A'.