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

Explanation

To find the coefficient of \( x^{-25} \) in the expansion of \( \left( \frac{x^2}{2} - \frac{3}{x^3} \right)^{15} \), we will use the binomial theorem.

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term in the expansion of \( (p + q)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} p^{n-r} q^r \]
Here, \( p = \frac{x^2}{2} \), \( q = -\frac{3}{x^3} \), and \( n = 15 \).

2. **Write the General Term**:
Substituting the values into the formula, we get:
\[ T_{r+1} = \binom{15}{r} \left( \frac{x^2}{2} \right)^{15-r} \left( -\frac{3}{x^3} \right)^r \]

3. **Simplify the General Term**:
This can be simplified as follows:
\[ T_{r+1} = \binom{15}{r} \left( \frac{x^{2(15-r)}}{2^{15-r}} \right) \left( -\frac{3^r}{x^{3r}} \right) \]
\[ = \binom{15}{r} (-3)^r \frac{x^{30 - 2r}}{2^{15 - r} x^{3r}} \]
\[ = \binom{15}{r} (-3)^r \frac{x^{30 - 5r}}{2^{15 - r}} \]

4. **Find the Power of \( x \)**:
We need to find the value of \( r \) such that the exponent of \( x \) is \( -25 \):
\[ 30 - 5r = -25 \]

5. **Solve for \( r \)**:
Rearranging the equation gives:
\[ 5r = 30 + 25 = 55 \quad \Rightarrow \quad r = \frac{55}{5} = 11 \]

6. **Substitute \( r \) back into the General Term**:
Now, substitute \( r = 11 \) into the general term:
\[ T_{12} = \binom{15}{11} (-3)^{11} \frac{1}{2^{15 - 11}} \]
\[ = \binom{15}{11} (-3)^{11} \frac{1}{2^4} \]

7. **Calculate \( \binom{15}{11} \)**:
\[ \binom{15}{11} = \binom{15}{4} = \frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1} = 1365 \]

8. **Calculate the Coefficient**:
Now, putting it all together:
\[ T_{12} = 1365 \cdot (-3)^{11} \cdot \frac{1}{16} \]
\[ = 1365 \cdot (-3)^{11} \cdot \frac{1}{16} \]

9. **Final Coefficient**:
The coefficient of \( x^{-25} \) is:
\[ = \frac{1365 \cdot (-3)^{11}}{16} \]