Let n is a rational number and x is a real number such that |x|lt1, then
`(1+x)^(n)=1+nx+(n(n-1)x^(2))/(2!)+(n(n-1)(n-2))/(3!).x^(3)+ . . .`
This can be used to find the sm of different series.
Q. 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. `((2)/(3))^(1//3)`

Answer

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 relate it to the binomial expansion.

### Step 1: Identify the pattern in the series

The series can be rewritten in a more recognizable form. The general term appears to be:

\[ T_k = \frac{1 \cdot 4 \cdot 7 \cdots (3k - 2)}{1 \cdot 2 \cdot 3 \cdots k} \cdot \frac{1}{3^{2k}} \]

This can be expressed using the Pochhammer symbol (or rising factorial):

\[ T_k = \frac{(3k-2)!!}{k!} \cdot \frac{1}{3^{2k}} \]

where \( (3k-2)!! \) is the double factorial.

### Step 2: Relate to the Binomial Theorem

We can express the series in terms of the binomial expansion. The binomial theorem states that:

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

For our case, we can relate \( n \) to the coefficients in our series.

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

From the series, we can see that:

- \( n = -\frac{1}{3} \)
- \( x = \frac{1}{3} \)

### Step 4: Use the Binomial Expansion

Using the binomial series expansion for \( (1+x)^n \):

\[ (1 + \frac{1}{3})^{-\frac{1}{3}} = \left(\frac{4}{3}\right)^{-\frac{1}{3}} \]

### Step 5: Calculate the result

Now we calculate:

\[ \left(\frac{4}{3}\right)^{-\frac{1}{3}} = \frac{3^{\frac{1}{3}}}{4^{\frac{1}{3}}} = \frac{3^{\frac{1}{3}}}{2^{\frac{2}{3}}} \]

### Final Result

Thus, the sum of the series is:

\[ S = \frac{3^{\frac{1}{3}}}{2^{\frac{2}{3}}} \]