The value of the product `6^((1)/(2))xx6^((1)/(4))xx6^((1)/(8))xx6^((1)/(16))xx....` up to infinite terms is

Options

A. 6
B. 36
C. 216
D. 512

Explanation

To find the value of the infinite product \( 6^{\frac{1}{2}} \times 6^{\frac{1}{4}} \times 6^{\frac{1}{8}} \times 6^{\frac{1}{16}} \times \ldots \), we can simplify the expression step by step.

### Step 1: Rewrite the Product
We can express the product in terms of exponents:
\[ P = 6^{\frac{1}{2}} \times 6^{\frac{1}{4}} \times 6^{\frac{1}{8}} \times 6^{\frac{1}{16}} \times \ldots \]
Using the property of exponents \( a^m \times a^n = a^{m+n} \), we can combine the exponents:
\[ P = 6^{\left(\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \ldots\right)} \]

### Step 2: Identify the Series
Now we need to find the sum of the series:
\[ S = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \ldots \]
This series is a geometric series where the first term \( a = \frac{1}{2} \) and the common ratio \( r = \frac{1}{2} \).

### Step 3: Sum of the Infinite Geometric Series
The sum \( S \) of an infinite geometric series can be calculated using the formula:
\[ S = \frac{a}{1 - r} \]
Substituting the values:
\[ S = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1 \]

### Step 4: Substitute Back into the Product
Now that we have the sum \( S = 1 \), we can substitute it back into our expression for \( P \):
\[ P = 6^{S} = 6^{1} = 6 \]

### Final Answer
Thus, the value of the product is:
\[ \boxed{6} \]