The value of the product: `6^((1)/(2))xx6^(1/4)xx6^(1/8)xx6^(1/(16))xx`. . . Up to infinite terms is:
(a)6
(b)36
(c)216
(d)512

Options

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

Answer

To solve the problem, we need to evaluate the infinite product:

\[ 6^{\frac{1}{2}} \times 6^{\frac{1}{4}} \times 6^{\frac{1}{8}} \times 6^{\frac{1}{16}} \times \ldots \]

### Step 1: Combine the exponents
Since the bases are the same (all are 6), we can combine the exponents by adding them together:

\[ 6^{\left(\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \ldots\right)} \]

### Step 2: Identify the series
The series in the exponent is:

\[ \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \ldots \]

This is a geometric series where the first term \( a = \frac{1}{2} \) and the common ratio \( r = \frac{1}{2} \).

### Step 3: Use the formula for the sum of an 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 of \( a \) and \( r \):

\[ S = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1 \]

### Step 4: Substitute the sum back into the exponent
Now we can substitute the sum back into the exponent:

\[ 6^{1} \]

### Step 5: Final result
Thus, the value of the product is:

\[ 6^{1} = 6 \]

### Conclusion
The final answer is:

\[ \text{(a) } 6 \]