The value of `9^(1//3)xx9^(1//9)xx9^(1//27)xx……` to `oo` is

Options

A. 1
B. 3
C. 9
D. None of these

Answer

To solve the problem \( 9^{\frac{1}{3}} \times 9^{\frac{1}{9}} \times 9^{\frac{1}{27}} \times \ldots \) to infinity, we can follow these steps:

### Step 1: Identify the series
The expression can be rewritten using the properties of exponents:
\[ 9^{\frac{1}{3}} \times 9^{\frac{1}{9}} \times 9^{\frac{1}{27}} \times \ldots = 9^{\left(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots\right)} \]

### Step 2: Recognize the series as a geometric series
The series inside the exponent is:
\[ \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots \]
This is a geometric series where the first term \( a = \frac{1}{3} \) and the common ratio \( r = \frac{1}{3} \).

### 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 \( r \):
\[ S = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2} \]

### Step 4: Substitute back into the exponent
Now we substitute this sum back into the exponent:
\[ 9^{\left(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots\right)} = 9^{\frac{1}{2}} \]

### Step 5: Simplify the expression
Since \( 9 = 3^2 \), we can rewrite:
\[ 9^{\frac{1}{2}} = (3^2)^{\frac{1}{2}} = 3^{2 \cdot \frac{1}{2}} = 3^1 = 3 \]

### Final Answer
Thus, the value of the infinite product \( 9^{\frac{1}{3}} \times 9^{\frac{1}{9}} \times 9^{\frac{1}{27}} \times \ldots \) is \( \boxed{3} \).

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