the coefficient of `x^(r)` in the expansion of
`(1 - 4x )^(-1//2)`, is

Options

A. `((2r)1)/((r !)2^(r))`
B. `""^(2r)C_(r)`
C. `(1.35….. (2r -1))/(2^(r) r!)`
D. none of these

Explanation

To find the coefficient of \( x^r \) in the expansion of \( (1 - 4x)^{-1/2} \), we can use the Binomial Theorem for negative exponents. The general term in the expansion of \( (1 - y)^{-n} \) is given by:

\[ T_{r+1} = \frac{n(n+1)(n+2)\cdots(n+r-1)}{r!} y^r \]

For our case, we have \( n = -\frac{1}{2} \) and \( y = 4x \). Thus, we can write:

\[ T_{r+1} = \frac{-\frac{1}{2} \left(-\frac{1}{2} + 1\right) \left(-\frac{1}{2} + 2\right) \cdots \left(-\frac{1}{2} + r - 1\right)}{r!} (4x)^r \]

Now, let's simplify this step by step.

### Step 1: Substitute \( n \) and \( y \) into the general term

Substituting \( n = -\frac{1}{2} \) and \( y = 4x \):

\[ T_{r+1} = \frac{-\frac{1}{2} \left(\frac{1}{2}\right) \left(\frac{3}{2}\right) \cdots \left(-\frac{1}{2} + r - 1\right)}{r!} (4x)^r \]

### Step 2: Write the product in the numerator

The numerator becomes:

\[ -\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{3}{2} \cdots \left(-\frac{1}{2} + r - 1\right) = \frac{(-1)^r}{2^r} \cdot (1)(3)(5)\cdots(2r-1) \]

### Step 3: Express the product of odd numbers

The product of the first \( r \) odd numbers can be expressed as:

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

### Step 4: Combine everything

Now we can combine everything into our term:

\[ T_{r+1} = \frac{(-1)^r}{2^r} \cdot \frac{(2r)!}{2^r r!} \cdot (4x)^r \]

This simplifies to:

\[ T_{r+1} = \frac{(-1)^r (2r)!}{(2^r r!)^2} \cdot 4^r x^r \]

### Step 5: Simplify the coefficient

Now, since \( 4^r = (2^2)^r = 2^{2r} \):

\[ T_{r+1} = \frac{(-1)^r (2r)!}{(2^r r!)^2} \cdot 2^{2r} x^r = \frac{(-1)^r (2r)!}{r! r!} x^r \]

### Step 6: Final coefficient

Thus, the coefficient of \( x^r \) is:

\[ \frac{(-1)^r (2r)!}{(r!)^2} \]

This is the required coefficient of \( x^r \) in the expansion of \( (1 - 4x)^{-1/2} \).