`2.4^((-1)/(4))*8^((1)/(9))*16^((-1)/(16))*32^((1)/(25))*64^((-1)/(36))*……….oo=`

Options

A. `3^(log3)`
B. `3^(log2)`
C. `2^(log2)`
D. `2^(log3)`

Answer

To solve the given expression:

\[ 2 \cdot 4^{-\frac{1}{4}} \cdot 8^{\frac{1}{9}} \cdot 16^{-\frac{1}{16}} \cdot 32^{\frac{1}{25}} \cdot 64^{-\frac{1}{36}} \cdots \]

we will break down each term and simplify it step by step.

### Step 1: Rewrite the terms in powers of 2

We know that:
- \( 4 = 2^2 \)
- \( 8 = 2^3 \)
- \( 16 = 2^4 \)
- \( 32 = 2^5 \)
- \( 64 = 2^6 \)

Thus, we can rewrite the expression as:

\[ 2 \cdot (2^2)^{-\frac{1}{4}} \cdot (2^3)^{\frac{1}{9}} \cdot (2^4)^{-\frac{1}{16}} \cdot (2^5)^{\frac{1}{25}} \cdot (2^6)^{-\frac{1}{36}} \cdots \]

### Step 2: Simplify the powers

Using the property of exponents \( (a^m)^n = a^{m \cdot n} \), we can simplify each term:

\[ = 2 \cdot 2^{-\frac{2}{4}} \cdot 2^{\frac{3}{9}} \cdot 2^{-\frac{4}{16}} \cdot 2^{\frac{5}{25}} \cdot 2^{-\frac{6}{36}} \cdots \]

This simplifies to:

\[ = 2^{1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots} \]

### Step 3: Identify the series

The exponent is a series:

\[ 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots \]

This series can be recognized as the alternating harmonic series, which converges to \( \log(2) \).

### Step 4: Substitute back into the expression

Now we can substitute back into our expression:

\[ = 2^{\log(2)} \]

### Step 5: Simplify further

Using the property \( a^{\log_a(b)} = b \), we have:

\[ = 2^{\log_2(2)} = 2 \]

### Final Answer

Thus, the value of the given expression is:

\[ \boxed{2} \]