The general term of `(2a - 3b)^(-1//2)` is

Options

A. `(1.3.5……(2r - 5))/(r!) (1)/(sqrt(2a)) ((3b)/(4a))^r`
B. `(1.3.5……(2r - 3))/(r!) (1)/(sqrt(2a)) ((3b)/(4a))^r`
C. `(1.3.5……(2r - 1))/(r!) (1)/(sqrt(2a)) ((3b)/(4a))^r`
D. `(1.3.5……(2r - 3))/(r!) (1)/(sqrta) ((3b)/(4a))^r`

Answer

To find the general term of the expression \((2a - 3b)^{-1/2}\), we can use the Binomial Theorem for negative and fractional exponents. The Binomial Theorem states that:

\[ (x + y)^n = \sum_{r=0}^{\infty} \binom{n}{r} x^{n-r} y^r \]

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

### Step 1: Identify the General Term
The general term \(T_{r+1}\) in the expansion of \((x + y)^n\) is given by:

\[ T_{r+1} = \binom{n}{r} x^{n-r} y^r \]

Substituting our values, we have:

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

### Step 2: Simplify the General Term
Next, we simplify the general term:

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

Using the property of binomial coefficients for negative integers:

\[ \binom{n}{r} = \frac{(-1)^r (-n)(-n-1)(-n-2)\cdots(-n-r+1)}{r!} = (-1)^r \frac{(-1/2)(-3/2)(-5/2)\cdots(-\frac{1}{2} - (r-1))}{r!} \]

### Step 3: Calculate the Coefficient
The coefficient can be calculated as:

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

### Step 4: Combine Terms
Now, we can combine the terms:

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

This simplifies to:

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

### Step 5: Final Form of the General Term
Thus, the general term can be expressed as:

\[ T_{r+1} = \frac{(-1)^r \cdot (2r - 1)!!}{2^r \cdot r!} \cdot (2a)^{-\frac{1}{2} - r} \cdot (3b)^r \]

Where \((2r - 1)!!\) denotes the double factorial of odd numbers.

### Final Answer
The general term of \((2a - 3b)^{-1/2}\) is:

\[ T_{r+1} = \frac{(-1)^r \cdot (2r - 1)!!}{2^r \cdot r!} \cdot (2a)^{-\frac{1}{2} - r} \cdot (3b)^r \]