`2^(1//4).4^(1//8).8^(1//16).16^(1//32)….` is equal to

Options

A. 2
B. `2^(2)`
C. `sqrt(2)`
D. `2^(1//4)`

Answer

To solve the expression \(2^{\frac{1}{4}} \cdot 4^{\frac{1}{8}} \cdot 8^{\frac{1}{16}} \cdot 16^{\frac{1}{32}} \cdots\), we will first rewrite each term in the product in terms of base 2.

### Step 1: Rewrite each term
- \(4^{\frac{1}{8}} = (2^2)^{\frac{1}{8}} = 2^{\frac{2}{8}} = 2^{\frac{1}{4}}\)
- \(8^{\frac{1}{16}} = (2^3)^{\frac{1}{16}} = 2^{\frac{3}{16}}\)
- \(16^{\frac{1}{32}} = (2^4)^{\frac{1}{32}} = 2^{\frac{4}{32}} = 2^{\frac{1}{8}}\)

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

### Step 2: Combine the powers of 2
The product can be rewritten as:
\[ 2^{\left(\frac{1}{4} + \frac{1}{4} + \frac{3}{16} + \frac{1}{8} + \cdots\right)} \]

### Step 3: Identify the series
Now, we need to find the sum of the series:
\[ S = \frac{1}{4} + \frac{1}{4} + \frac{3}{16} + \frac{1}{8} + \cdots \]

### Step 4: Rewrite the series
Notice that:
- The first term is \(\frac{1}{4}\).
- The second term is also \(\frac{1}{4}\).
- The third term can be rewritten as \(\frac{3}{16} = \frac{3}{2^4}\).
- The fourth term can be rewritten as \(\frac{1}{8} = \frac{1}{2^3}\).

The series can be expressed as:
\[ S = \sum_{n=1}^{\infty} \frac{n}{2^{n+1}} \]

### Step 5: Use the formula for the sum of the series
To find \(S\), we can use the formula for the sum of an infinite series:
\[ S = \sum_{n=1}^{\infty} \frac{n}{r^n} = \frac{r}{(1-r)^2} \text{ for } |r| < 1 \]
Here, \(r = \frac{1}{2}\):
\[ S = \frac{\frac{1}{2}}{(1 - \frac{1}{2})^2} = \frac{\frac{1}{2}}{\left(\frac{1}{2}\right)^2} = \frac{\frac{1}{2}}{\frac{1}{4}} = 2 \]

### Step 6: Substitute back into the expression
Now substituting back into our expression:
\[ P = 2^S = 2^2 = 4 \]

### Step 7: Conclusion
Thus, the value of the original expression \(2^{\frac{1}{4}} \cdot 4^{\frac{1}{8}} \cdot 8^{\frac{1}{16}} \cdots\) is equal to \(4\).