Find the coefficient of `x^(15)` in the expansion of `(x-x^2)^(10)`.

Answer

To find the coefficient of \( x^{15} \) in the expansion of \( (x - x^2)^{10} \), we can follow these steps:

### Step 1: Understand the Binomial Expansion
The expression \( (x - x^2)^{10} \) can be expanded using the Binomial Theorem, which states that:
\[ (a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r \]
In our case, \( a = x \), \( b = -x^2 \), and \( n = 10 \).

### Step 2: Write the General Term
The general term \( T_{r+1} \) in the expansion is given by:
\[ T_{r+1} = \binom{10}{r} (x)^{10-r} (-x^2)^r \]
This simplifies to:
\[ T_{r+1} = \binom{10}{r} x^{10-r} (-1)^r x^{2r} = \binom{10}{r} (-1)^r x^{10 + r} \]

### Step 3: Set Up the Equation for \( x^{15} \)
We need to find the term where the power of \( x \) is 15:
\[ 10 + r = 15 \]
Solving for \( r \):
\[ r = 15 - 10 = 5 \]

### Step 4: Find the Coefficient for \( r = 5 \)
Now, substitute \( r = 5 \) into the general term:
\[ T_{6} = \binom{10}{5} (-1)^5 x^{15} \]
The coefficient of \( x^{15} \) is:
\[ \binom{10}{5} (-1)^5 \]

### Step 5: Calculate \( \binom{10}{5} \)
We calculate \( \binom{10}{5} \):
\[ \binom{10}{5} = \frac{10!}{5! \cdot 5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252 \]

### Step 6: Determine the Final Coefficient
Since \( (-1)^5 = -1 \), the coefficient of \( x^{15} \) is:
\[ 252 \times (-1) = -252 \]

### Final Answer
The coefficient of \( x^{15} \) in the expansion of \( (x - x^2)^{10} \) is \( \boxed{-252} \).

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