Find the coefficient of `x^(7)` in the expansion of `(2x^(2)-(1)/(x))^(20)`

Explanation

To find the coefficient of \( x^7 \) in the expansion of \( (2x^2 - \frac{1}{x})^{20} \), we can use the Binomial Theorem.

### Step-by-Step Solution:

1. **Identify the Binomial Expansion**:
The expression can be written in the form \( (p + q)^n \) where:
- \( p = 2x^2 \)
- \( q = -\frac{1}{x} \)
- \( n = 20 \)

2. **General Term of the Expansion**:
The general term \( T_{r+1} \) in the expansion of \( (p + q)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} p^{n-r} q^r \]
Substituting the values of \( p \), \( q \), and \( n \):
\[ T_{r+1} = \binom{20}{r} (2x^2)^{20-r} \left(-\frac{1}{x}\right)^r \]

3. **Simplifying the General Term**:
\[ T_{r+1} = \binom{20}{r} (2^{20-r} (x^2)^{20-r}) \left(-1\right)^r \left(\frac{1}{x^r}\right) \]
This simplifies to:
\[ T_{r+1} = \binom{20}{r} (-1)^r 2^{20-r} x^{40 - 2r - r} = \binom{20}{r} (-1)^r 2^{20-r} x^{40 - 3r} \]

4. **Finding the Coefficient of \( x^7 \)**:
We need to find the value of \( r \) such that the exponent of \( x \) is 7:
\[ 40 - 3r = 7 \]
Rearranging gives:
\[ 3r = 40 - 7 = 33 \quad \Rightarrow \quad r = \frac{33}{3} = 11 \]

5. **Substituting \( r \) back into the General Term**:
Now, substitute \( r = 11 \) into the general term:
\[ T_{12} = \binom{20}{11} (-1)^{11} 2^{20-11} x^{40 - 3 \cdot 11} \]
This simplifies to:
\[ T_{12} = \binom{20}{11} (-1)^{11} 2^9 x^7 \]

6. **Finding the Coefficient**:
The coefficient of \( x^7 \) is:
\[ \text{Coefficient} = \binom{20}{11} (-1)^{11} 2^9 \]
Since \( (-1)^{11} = -1 \), we have:
\[ \text{Coefficient} = -\binom{20}{11} \cdot 2^9 \]

### Final Answer:
The coefficient of \( x^7 \) in the expansion of \( (2x^2 - \frac{1}{x})^{20} \) is:
\[ -\binom{20}{11} \cdot 2^9 \]