Find the coefficient of `x^(-25)` in the expansion of `((x^(2))/(2)-(3)/(x^(3)))^(15)`

Options

A. `(-1365)/(16)xx3^(11)`
B. `(1365)/(16)xx3^(11)`
C. `(-16)/(1365)xx3^(11)`
D. None of these

Explanation

To find the coefficient of \( x^{-25} \) in the expansion of \( \left( \frac{x^2}{2} - \frac{3}{x^3} \right)^{15} \), we can follow these steps:

### Step 1: Write the general term of the expansion
The general term (or \( (r+1)^{th} \) term) in the binomial expansion is given by:

\[ T_{r+1} = \binom{15}{r} \left( \frac{x^2}{2} \right)^r \left( -\frac{3}{x^3} \right)^{15-r} \]

### Step 2: Simplify the general term
Now, simplifying the general term, we have:

\[ T_{r+1} = \binom{15}{r} \left( \frac{x^2}{2} \right)^r \left( -3 \right)^{15-r} \left( \frac{1}{x^3} \right)^{15-r} \]

This simplifies to:

\[ T_{r+1} = \binom{15}{r} \cdot \frac{(-3)^{15-r}}{2^r} \cdot x^{2r} \cdot x^{-3(15-r)} \]

### Step 3: Combine the powers of \( x \)
Combining the powers of \( x \):

\[ T_{r+1} = \binom{15}{r} \cdot \frac{(-3)^{15-r}}{2^r} \cdot x^{2r - 45 + 3r} \]

This simplifies to:

\[ T_{r+1} = \binom{15}{r} \cdot \frac{(-3)^{15-r}}{2^r} \cdot x^{5r - 45} \]

### Step 4: Set the exponent of \( x \) to \(-25\)
We need the exponent of \( x \) to equal \(-25\):

\[ 5r - 45 = -25 \]

### Step 5: Solve for \( r \)
Solving for \( r \):

\[ 5r = 45 - 25 \]
\[ 5r = 20 \implies r = 4 \]

### Step 6: Substitute \( r \) back into the general term
Now, substitute \( r = 4 \) back into the expression for the general term:

\[ T_{5} = \binom{15}{4} \cdot \frac{(-3)^{15-4}}{2^4} \]

### Step 7: Calculate the coefficient
Calculating the coefficient:

\[ T_{5} = \binom{15}{4} \cdot \frac{(-3)^{11}}{16} \]

Calculating \( \binom{15}{4} \):

\[ \binom{15}{4} = \frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1} = 1365 \]

Thus,

\[ T_{5} = 1365 \cdot \frac{(-3)^{11}}{16} \]

### Step 8: Final coefficient
The coefficient of \( x^{-25} \) is:

\[ \text{Coefficient} = \frac{-1365 \cdot 3^{11}}{16} \]