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 follow these steps:

### Step 1: Write the general term of the expansion
The general term (r+1) in the binomial expansion of \( (a + b)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
In our case, \( a = \frac{x^2}{2} \), \( b = -\frac{3}{x^3} \), and \( n = 15 \). Thus, the general term becomes:
\[ T_{r+1} = \binom{15}{r} \left( \frac{x^2}{2} \right)^{15-r} \left( -\frac{3}{x^3} \right)^r \]

### Step 2: Simplify the general term
Now, we simplify \( T_{r+1} \):
\[ T_{r+1} = \binom{15}{r} \left( \frac{x^{2(15-r)}}{2^{15-r}} \right) \left( -\frac{3^r}{x^{3r}} \right) \]
This simplifies to:
\[ T_{r+1} = \binom{15}{r} \cdot \frac{(-3)^r}{2^{15-r}} \cdot x^{2(15-r) - 3r} \]
\[ = \binom{15}{r} \cdot \frac{(-3)^r}{2^{15-r}} \cdot x^{30 - 2r - 3r} \]
\[ = \binom{15}{r} \cdot \frac{(-3)^r}{2^{15-r}} \cdot x^{30 - 5r} \]

### Step 3: Set the exponent of \( x \) to -25
We need the exponent of \( x \) to be -25:
\[ 30 - 5r = -25 \]
Solving for \( r \):
\[ 30 + 25 = 5r \implies 55 = 5r \implies r = 11 \]

### Step 4: Substitute \( r \) back to find the coefficient
Now we substitute \( r = 11 \) back into the expression for the general term:
\[ T_{12} = \binom{15}{11} \cdot \frac{(-3)^{11}}{2^{15-11}} \cdot x^{-25} \]
Calculating \( \binom{15}{11} \):
\[ \binom{15}{11} = \binom{15}{4} = \frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1} = 1365 \]

### Step 5: Calculate the coefficient
Now substituting back:
\[ T_{12} = 1365 \cdot \frac{(-3)^{11}}{2^4} \]
Calculating \( 2^4 = 16 \):
\[ T_{12} = 1365 \cdot \frac{-3^{11}}{16} \]

### Final Result
Thus, the coefficient of \( x^{-25} \) is:
\[ \text{Coefficient} = \frac{-1365 \cdot 3^{11}}{16} \]