Let `P = 3^(1//3). 3^(2//9) . 3^(3//27)…oo`, then `P^(1//3)` is equal to

Options

A. `3^(2//3)`
B. `sqrt(3)`
C. `3^(1//3)`
D. `3^(1//4)`

Answer

To solve the problem, we need to evaluate the infinite product \( P = 3^{1/3} \cdot 3^{2/9} \cdot 3^{3/27} \cdots \) and then find \( P^{1/3} \).

### Step-by-Step Solution:

1. **Express the Product**:
We can express \( P \) as:
\[ P = 3^{1/3} \cdot 3^{2/9} \cdot 3^{3/27} \cdots \]
This can be rewritten using the property of exponents:
\[ P = 3^{\left(\frac{1}{3} + \frac{2}{9} + \frac{3}{27} + \cdots \right)} \]

2. **Identify the Series**:
The series in the exponent is:
\[ S = \frac{1}{3} + \frac{2}{9} + \frac{3}{27} + \cdots \]
We can rewrite the terms:
\[ S = \frac{1}{3} + \frac{2}{3^2} + \frac{3}{3^3} + \cdots \]

3. **Multiply by \( \frac{1}{3} \)**:
To find a relation for \( S \), multiply the entire series by \( \frac{1}{3} \):
\[ \frac{S}{3} = \frac{1}{3^2} + \frac{2}{3^3} + \frac{3}{3^4} + \cdots \]

4. **Subtract the Two Series**:
Now, subtract \( \frac{S}{3} \) from \( S \):
\[ S - \frac{S}{3} = \left(\frac{1}{3} - \frac{1}{3^2}\right) + \left(\frac{2}{9} - \frac{2}{27}\right) + \left(\frac{3}{27} - \frac{3}{81}\right) + \cdots \]
This simplifies to:
\[ \frac{2S}{3} = \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \cdots \]

5. **Recognize the Geometric Series**:
The right-hand side is a geometric series with first term \( \frac{1}{3} \) and common ratio \( \frac{1}{3} \):
\[ \frac{1}{3} \cdot \frac{1}{1 - \frac{1}{3}} = \frac{1/3}{2/3} = \frac{1}{2} \]

6. **Solve for \( S \)**:
Thus, we have:
\[ \frac{2S}{3} = \frac{1}{2} \]
Multiplying both sides by \( \frac{3}{2} \):
\[ S = \frac{3}{4} \]

7. **Substitute Back into \( P \)**:
Now substituting \( S \) back into the expression for \( P \):
\[ P = 3^{S} = 3^{3/4} \]

8. **Find \( P^{1/3} \)**:
Finally, we need to find \( P^{1/3} \):
\[ P^{1/3} = \left(3^{3/4}\right)^{1/3} = 3^{3/4 \cdot 1/3} = 3^{1/4} \]

### Final Answer:
Thus, \( P^{1/3} = 3^{1/4} \).