The sum of the series
`1 + (1)/(3^(2)) + (1 *4)/(1*2) (1)/(3^(4))+( 1 * 4 * 7)/(1 *2*3)(1)/(3^(6)) + ..., ` is

Options

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

Explanation

To find the sum of the series

\[ S = 1 + \frac{1}{3^2} + \frac{1 \cdot 4}{1 \cdot 2} \cdot \frac{1}{3^4} + \frac{1 \cdot 4 \cdot 7}{1 \cdot 2 \cdot 3} \cdot \frac{1}{3^6} + \ldots \]

we can observe that the series can be expressed in terms of a generalized binomial expansion.

### Step 1: Identify the general term of the series

The general term of the series can be expressed as:

\[ T_n = \frac{(3n-2)(3n-5)\ldots(1)}{n!} \cdot \frac{1}{3^{2n}} \]

for \( n = 0, 1, 2, \ldots \)

### Step 2: Recognize the pattern

The numerator of the general term can be recognized as a double factorial, which can also be expressed in terms of the Gamma function or Pochhammer symbols. The series resembles the expansion of \( (1-x)^{-k} \).

### Step 3: Use the binomial series expansion

The binomial series expansion for \( (1-x)^{-k} \) is given by:

\[ (1-x)^{-k} = \sum_{n=0}^{\infty} \frac{(k)_n}{n!} x^n \]

where \( (k)_n \) is the Pochhammer symbol (rising factorial).

### Step 4: Set up the equation

In our case, we can set \( x = \frac{1}{9} \) and \( k = \frac{1}{3} \). Thus, we can write:

\[ S = (1 - \frac{1}{9})^{-1/3} \]

### Step 5: Simplify the expression

Now we simplify:

\[ S = \left(\frac{8}{9}\right)^{-1/3} = \left(\frac{9}{8}\right)^{1/3} \]

### Step 6: Final result

Thus, the sum of the series is:

\[ S = \left(\frac{9}{8}\right)^{1/3} \]