The value of `2^(1/4).4^(1/8).8^(1/16),,,,,,,oo` is equal to.

Options

A. 4
B. `sqrt(2)`
C. `8
D. 2

Answer

To find the value of the infinite product \( 2^{1/4} \cdot 4^{1/8} \cdot 8^{1/16} \cdots \), we can follow these steps:

### Step 1: Rewrite the terms in the product
We start by rewriting \( 4 \) and \( 8 \) in terms of base \( 2 \):
- \( 4 = 2^2 \)
- \( 8 = 2^3 \)

Thus, we can express the product as:
\[ 2^{1/4} \cdot (2^2)^{1/8} \cdot (2^3)^{1/16} \cdots \]

### Step 2: Simplify the expression
Now, we can simplify each term:
\[ 2^{1/4} \cdot 2^{2/8} \cdot 2^{3/16} = 2^{1/4} \cdot 2^{1/4} \cdot 2^{3/16} \]
This gives us:
\[ 2^{1/4 + 1/4 + 3/16} \]

### Step 3: Find a general term for the series
We can observe that the exponents form a series:
\[ \frac{1}{4} + \frac{2}{8} + \frac{3}{16} + \cdots \]
This can be rewritten as:
\[ \sum_{n=1}^{\infty} \frac{n}{2^{n+1}} \]

### Step 4: Calculate the sum of the series
To find the sum \( S = \sum_{n=1}^{\infty} \frac{n}{2^{n+1}} \), we can use the formula for the sum of an infinite series:
\[ S = \frac{a}{(1 - r)^2} \]
where \( a \) is the first term and \( r \) is the common ratio. Here, \( a = \frac{1}{2} \) (the first term when \( n=1 \)) and \( r = \frac{1}{2} \).

Calculating:
\[ S = \frac{\frac{1}{2}}{(1 - \frac{1}{2})^2} = \frac{\frac{1}{2}}{\left(\frac{1}{2}\right)^2} = \frac{\frac{1}{2}}{\frac{1}{4}} = 2 \]

### Step 5: Substitute back into the exponent
Now we can substitute back into our expression for the product:
\[ 2^{S} = 2^{2} = 4 \]

### Final Answer
Thus, the value of \( 2^{1/4} \cdot 4^{1/8} \cdot 8^{1/16} \cdots \) is:
\[ \boxed{4} \]