Show that ` 1 + 1/2.(3/5) + (1.3)/(2.4) (3/5)^2 + (1.3.5)/(2.4.6) (3/5)^3 + …… = sqrt([5/2])`

Answer

To show that

\[ 1 + \frac{1}{2} \cdot \frac{3}{5} + \frac{1 \cdot 3}{2 \cdot 4} \cdot \left(\frac{3}{5}\right)^2 + \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} \cdot \left(\frac{3}{5}\right)^3 + \ldots = \sqrt{\frac{5}{2}}, \]

we will analyze the left-hand side and relate it to a known series.

### Step 1: Identify the general term

The series can be expressed in terms of a general term. The \( n \)-th term can be written as:

\[ T_n = \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)} \cdot \left(\frac{3}{5}\right)^n. \]

This can also be expressed using double factorials:

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

### Step 2: Recognize the series as a binomial expansion

The series resembles the binomial expansion of \( (1 + x)^n \) where \( n \) is a negative half-integer. Specifically, we can relate it to:

\[ (1-x)^{-1/2} = \sum_{n=0}^{\infty} \frac{(2n)!}{(n!)^2 (2^n)} x^n. \]

### Step 3: Set up the equation

By substituting \( x = \frac{3}{5} \) into the binomial expansion, we have:

\[ (1 - \frac{3}{5})^{-1/2} = (1 - 0.6)^{-1/2} = (0.4)^{-1/2} = \frac{1}{\sqrt{0.4}} = \frac{1}{\sqrt{\frac{2}{5}}} = \sqrt{\frac{5}{2}}. \]

### Step 4: Conclusion

Thus, we conclude that:

\[ 1 + \frac{1}{2} \cdot \frac{3}{5} + \frac{1 \cdot 3}{2 \cdot 4} \cdot \left(\frac{3}{5}\right)^2 + \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} \cdot \left(\frac{3}{5}\right)^3 + \ldots = \sqrt{\frac{5}{2}}. \]