The value of `3^(1//3) . 9^(1//18) . 27^(1//81)` …. ∞

Options

A. a)3
B. b)9
C. c)27
D. d)none of these

Answer

To solve the expression \(3^{1/3} \cdot 9^{1/18} \cdot 27^{1/81} \cdots\) to infinity, we can follow these steps:

### Step 1: Rewrite the terms in the expression
We start by rewriting \(9\) and \(27\) in terms of base \(3\):
- \(9 = 3^2\)
- \(27 = 3^3\)

Thus, we can rewrite the expression as:
\[ 3^{1/3} \cdot (3^2)^{1/18} \cdot (3^3)^{1/81} \]

### Step 2: Simplify the expression
Now, we simplify each term:
- The first term remains \(3^{1/3}\).
- The second term becomes \(3^{2/18} = 3^{1/9}\).
- The third term becomes \(3^{3/81} = 3^{1/27}\).

So, the expression now looks like:
\[ 3^{1/3} \cdot 3^{1/9} \cdot 3^{1/27} \cdots \]

### Step 3: Combine the exponents
Since the bases are the same, we can add the exponents:
\[ 3^{(1/3) + (1/9) + (1/27) + \cdots} \]

### Step 4: Identify the series
The series in the exponent is:
\[ \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \cdots \]
This is a geometric series where:
- The first term \(a = \frac{1}{3}\)
- The common ratio \(r = \frac{1}{3}\)

### Step 5: Calculate the sum of the infinite series
The sum \(S\) of an infinite geometric series can be calculated using the formula:
\[ S = \frac{a}{1 - r} \]
Substituting the values:
\[ S = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2} \]

### Step 6: Substitute back into the expression
Now substituting back into the expression, we have:
\[ 3^{(1/2)} = \sqrt{3} \]

### Final Answer
Thus, the value of the original expression \(3^{1/3} \cdot 9^{1/18} \cdot 27^{1/81} \cdots\) to infinity is:
\[ \sqrt{3} \]

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