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 sum of different series. Q. Sum of infinite series `1+(2)/(3)*(1)/(2)+(2)/(3)*(5)/(6)*(1)/(2^(2))+(2)/(3)*(5)/(6)*(8)/(9)*(1)/(2^(3))+ . . .oo` is

Options

A. (a)`2^(1//3)`
B. (b)`4^(1//3)`
C. (c)`8^(1//3)`
D. (d)`4^(2//3)`

Answer

To solve the infinite series given in the question, we can use the Binomial Theorem. The series we need to evaluate is:

\[ S = 1 + \frac{2}{3} \cdot \frac{1}{2} + \frac{2}{3} \cdot \frac{5}{6} \cdot \frac{1}{2^2} + \frac{2}{3} \cdot \frac{5}{6} \cdot \frac{8}{9} \cdot \frac{1}{2^3} + \ldots \]

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

We can observe that the series has a pattern. The first term is \(1\), the second term is \(\frac{2}{3} \cdot \frac{1}{2}\), the third term is \(\frac{2}{3} \cdot \frac{5}{6} \cdot \frac{1}{2^2}\), and so on.

The general term can be expressed as:

\[ T_n = \frac{2}{3} \cdot \frac{5}{6} \cdot \frac{8}{9} \cdots \cdot \frac{(3n-1)}{(3n)} \cdot \frac{1}{2^{n-1}} \]

### Step 2: Express the general term

The general term can be simplified. The product of the fractions can be expressed as:

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

### Step 3: Recognize the series as a binomial expansion

We can relate this series to the binomial expansion of \((1+x)^n\). Here, we can set \(x = \frac{1}{2}\) and \(n = -\frac{2}{3}\):

Using the Binomial Theorem:

\[ (1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \ldots \]

### Step 4: Substitute values into the binomial expansion

Substituting \(x = \frac{1}{2}\) and \(n = -\frac{2}{3}\):

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

### Step 5: Calculate the value

Now we need to calculate \(\left(\frac{3}{2}\right)^{-\frac{2}{3}}\):

\[ \left(\frac{3}{2}\right)^{-\frac{2}{3}} = \frac{1}{\left(\frac{3}{2}\right)^{\frac{2}{3}}} = \frac{1}{\left(\frac{3^2}{2^2}\right)^{\frac{1}{3}}} = \frac{1}{\left(\frac{9}{4}\right)^{\frac{1}{3}}} = \frac{2^{\frac{2}{3}}}{3^{\frac{2}{3}}} \]

### Final Answer

Thus, the sum of the infinite series is:

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