The sum of `1+1/4+(1.3)/(4.8)+(1.3.5)/(4.8.12)+.........oo` is

Answer

To find the sum of the series \( S = 1 + \frac{1 \cdot 3}{4 \cdot 8} + \frac{1 \cdot 3 \cdot 5}{4 \cdot 8 \cdot 12} + \ldots \), we can relate this series to the binomial theorem.

### Step-by-Step Solution:

1. **Identify the Pattern**:
The series can be expressed as:
\[ S = 1 + \frac{1 \cdot 3}{4 \cdot 8} + \frac{1 \cdot 3 \cdot 5}{4 \cdot 8 \cdot 12} + \ldots \]
The general term appears to be:
\[ T_n = \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{4^n \cdot n!} \]

2. **Recognize the Factorial Representation**:
The numerator \( 1 \cdot 3 \cdot 5 \cdots (2n-1) \) can be represented as \( \frac{(2n)!}{2^n n!} \). Thus, we can rewrite the general term:
\[ T_n = \frac{(2n)!}{2^n n! \cdot 4^n \cdot n!} = \frac{(2n)!}{(2^n \cdot 4^n) \cdot (n!)^2} = \frac{(2n)!}{(2^{2n}) \cdot (n!)^2} \]

3. **Use the Binomial Theorem**:
The series \( S \) can be recognized as the expansion of \( (1-x)^{-1/2} \) evaluated at \( x = \frac{1}{4} \):
\[ (1-x)^{-1/2} = \sum_{n=0}^{\infty} \frac{(2n)!}{(2^n \cdot n!)^2} x^n \]
Substituting \( x = \frac{1}{4} \):
\[ S = (1 - \frac{1}{4})^{-1/2} = \left(\frac{3}{4}\right)^{-1/2} = \frac{2}{\sqrt{3}} \]

4. **Final Result**:
Thus, the sum of the series is:
\[ S = \frac{2}{\sqrt{3}} \]