The sum of the series
` 3 /(4*8) -(3*5)/(4*8*12)+(3*5*7)/(4*8*12*16)-...` is

Options

A. ` sqrt(3/2)-3/4`
B. ` sqrt(4/3)-3/4`
C. ` sqrt(3/2)-1/4`
D. ` sqrt(2/3)-1/4`

Explanation

To find the sum of the series

\[ S = \frac{3}{4 \cdot 8} - \frac{3 \cdot 5}{4 \cdot 8 \cdot 12} + \frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12 \cdot 16} - \ldots \]

we can start by rewriting the terms in a more manageable form.

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

The series can be expressed in terms of factorials and powers. The numerator of the terms appears to be a product of odd numbers, while the denominator is a product of even numbers.

The general term can be written as:

\[ T_n = \frac{3 \cdot 5 \cdot 7 \cdots (2n + 1)}{4^n \cdot (2n)!!} \]

Where \( (2n)!! \) denotes the double factorial of \( 2n \).

### Step 2: Factor out constants

We can factor out constants from the series. Notice that:

\[ \frac{3}{4 \cdot 8} = \frac{3}{32} \]

Thus, we can rewrite the series as:

\[ S = \frac{3}{32} \left( 1 - \frac{5}{3} \cdot \frac{1}{4} + \frac{5 \cdot 7}{3 \cdot 2!} \cdot \frac{1}{16} - \ldots \right) \]

### Step 3: Recognize the series as a Taylor series

The series resembles the Taylor series expansion for \( (1-x)^{-k} \) for some \( k \). Specifically, we can relate it to the series expansion of \( (1-x)^{-3/2} \).

### Step 4: Use the formula for the sum of the series

The sum of the series can be calculated using the formula for the sum of the series:

\[ S = \frac{3}{32} \cdot (1 - x)^{-3/2} \text{ at } x = \frac{1}{4} \]

Substituting \( x = \frac{1}{4} \):

\[ S = \frac{3}{32} \cdot (1 - \frac{1}{4})^{-3/2} = \frac{3}{32} \cdot \left(\frac{3}{4}\right)^{-3/2} \]

### Step 5: Calculate the final value

Calculating \( \left(\frac{3}{4}\right)^{-3/2} \):

\[ \left(\frac{3}{4}\right)^{-3/2} = \left(\frac{4}{3}\right)^{3/2} = \frac{8}{3\sqrt{3}} \]

Thus,

\[ S = \frac{3}{32} \cdot \frac{8}{3\sqrt{3}} = \frac{8}{32\sqrt{3}} = \frac{1}{4\sqrt{3}} \]

### Final Answer:

The sum of the series is

\[ \frac{1}{4\sqrt{3}} \]