The general term in the expansion of
`( 1- 2x)^(3//4)` , is

Options

A. `(-3)/(2^(r)r!)x^(2)`
B. `(-3^(r))/(2^(r)r!)x^(r)`
C. `(-3^(r))/(2^(r)(2r)!)x^(r)`
D. none of these

Explanation

To find the general term in the expansion of \( (1 - 2x)^{\frac{3}{4}} \), we can use the Binomial Theorem. The Binomial Theorem states that:

\[ (a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r \]

For our case, we have \( a = 1 \), \( b = -2x \), and \( n = \frac{3}{4} \).

### Step 1: Identify the General Term
The general term \( T_{r+1} \) in the expansion can be expressed as:

\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]

Substituting the values we have:

\[ T_{r+1} = \binom{\frac{3}{4}}{r} (1)^{\frac{3}{4}-r} (-2x)^r \]

### Step 2: Simplify the General Term
Since \( (1)^{\frac{3}{4}-r} = 1 \), we can simplify the term:

\[ T_{r+1} = \binom{\frac{3}{4}}{r} (-2x)^r \]

### Step 3: Write the Binomial Coefficient
The binomial coefficient \( \binom{n}{r} \) for non-integer \( n \) is given by:

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

So, we have:

\[ \binom{\frac{3}{4}}{r} = \frac{\frac{3}{4} \left(\frac{3}{4} - 1\right) \left(\frac{3}{4} - 2\right) \cdots \left(\frac{3}{4} - (r-1)\right)}{r!} \]

### Step 4: Combine Terms
Now substituting this into our general term:

\[ T_{r+1} = \frac{\frac{3}{4} \left(-\frac{1}{4}\right) \left(-\frac{5}{4}\right) \cdots \left(\frac{3}{4} - (r-1)\right)}{r!} (-2x)^r \]

### Step 5: Final Expression
Thus, the general term in the expansion of \( (1 - 2x)^{\frac{3}{4}} \) is:

\[ T_{r+1} = \frac{\frac{3}{4} \left(-\frac{1}{4}\right) \left(-\frac{5}{4}\right) \cdots \left(\frac{3}{4} - (r-1)\right)}{r!} (-2)^r x^r \]

### Summary
The general term \( T_{r+1} \) in the expansion of \( (1 - 2x)^{\frac{3}{4}} \) can be expressed in the above form, which includes the binomial coefficient for non-integer \( n \).