In the expansion of `(x^(4)-(1)/(x^(3)))^(15)` the coefficient of `x^(39)` is

Options

A. 1365
B. `-1365`
C. 455
D. `-455`

Answer

To find the coefficient of \( x^{39} \) in the expansion of \( (x^4 - \frac{1}{x^3})^{15} \), we can follow these steps:

### Step 1: Identify the General Term
The general term \( T_{r+1} \) 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, \( a = x^4 \), \( b = -\frac{1}{x^3} \), and \( n = 15 \). Thus, the general term becomes:
\[ T_{r+1} = \binom{15}{r} (x^4)^{15-r} \left(-\frac{1}{x^3}\right)^r \]

### Step 2: Simplify the General Term
Now, simplifying \( T_{r+1} \):
\[ T_{r+1} = \binom{15}{r} (x^{4(15-r)}) \left(-1\right)^r \frac{1}{x^{3r}} = \binom{15}{r} (-1)^r x^{60 - 4r - 3r} = \binom{15}{r} (-1)^r x^{60 - 7r} \]

### Step 3: Set the Power of \( x \) Equal to 39
To find the coefficient of \( x^{39} \), we set the exponent equal to 39:
\[ 60 - 7r = 39 \]

### Step 4: Solve for \( r \)
Rearranging the equation:
\[ 60 - 39 = 7r \implies 21 = 7r \implies r = 3 \]

### Step 5: Substitute \( r \) Back into the General Term
Now, substitute \( r = 3 \) back into the general term to find the coefficient:
\[ T_{4} = \binom{15}{3} (-1)^3 \]

### Step 6: Calculate the Coefficient
Calculating \( \binom{15}{3} \):
\[ \binom{15}{3} = \frac{15!}{3!(15-3)!} = \frac{15 \times 14 \times 13}{3 \times 2 \times 1} = \frac{2730}{6} = 455 \]
Thus, the term becomes:
\[ T_{4} = 455 \cdot (-1) = -455 \]

### Final Answer
The coefficient of \( x^{39} \) in the expansion of \( (x^4 - \frac{1}{x^3})^{15} \) is:
\[ \boxed{-455} \]