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 (a) `((3)/(2))^((1)/(3))` (b) `((5)/(4))^((1)/(3))` (c) `((3)/(2))^((1)/(6))` (d) None of these

Options

A. `((3)/(2))^((1)/(3))`
B. `((5)/(4))^((1)/(3))`
C. `((3)/(2))^((1)/(6))`
D. None of these

Explanation

To find the sum of the series

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

we can analyze the pattern of the series. The series can be expressed in a general form where the \(n\)-th term can be represented as:

\[ T_n = \frac{(3n-2)(3n-5)(3n-8)\ldots}{n!} \cdot \frac{1}{3^{2n}} \]

This indicates that we can relate this series to the binomial theorem.

### Step 1: Identify the series structure

The first term is 1, the second term is \(\frac{1}{3^2}\), and the third term is \(\frac{1 \cdot 4}{1 \cdot 2} \cdot \frac{1}{3^4}\). The general term can be identified as:

\[ T_n = \frac{(3n-2)(3n-5)(3n-8)\ldots}{n!} \cdot \frac{1}{3^{2n}} \]

### Step 2: Relate to binomial expansion

We can relate this series to the binomial expansion of \((1-x)^{-k}\) for some \(k\). The series resembles the expansion of \((1-x)^{-n}\) where \(x = \frac{1}{3^2}\).

### Step 3: Define \(x\) and \(n\)

Let \(x = \frac{1}{9}\) (since \(3^2 = 9\)) and \(n = -\frac{1}{3}\). The binomial expansion gives:

\[ (1-x)^{-n} = \sum_{k=0}^{\infty} \binom{n+k-1}{k} x^k \]

### Step 4: Substitute \(x\) and \(n\)

Substituting \(x = \frac{1}{9}\) and \(n = -\frac{1}{3}\):

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

### Step 5: Simplify the expression

Now we can simplify:

\[ \left(\frac{8}{9}\right)^{\frac{1}{3}} = \frac{2}{3^{\frac{1}{3}}} \]

### Step 6: Final expression

We can express the sum of the series as:

\[ \left(\frac{3}{2}\right)^{\frac{1}{3}} \]

### Conclusion

Thus, the sum of the series is:

\[ \left(\frac{3}{2}\right)^{\frac{1}{3}} \]

This corresponds to option (a).

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