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 follow these steps:

### Step 1: Identify the General Term
The general term 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, \( a = 2x^2 \), \( b = -\frac{1}{x} \), and \( n = 20 \).

### Step 2: Write the General Term
Substituting the values into the general term formula, we have:
\[ T_{r+1} = \binom{20}{r} (2x^2)^{20-r} \left(-\frac{1}{x}\right)^r \]
This simplifies to:
\[ T_{r+1} = \binom{20}{r} (2^{20-r} x^{2(20-r)}) \left(-1\right)^r x^{-r} \]
Combining the powers of \( x \):
\[ T_{r+1} = \binom{20}{r} 2^{20-r} (-1)^r x^{40 - 3r} \]

### Step 3: Set Up the Equation for \( x^7 \)
We need to find the coefficient of \( x^7 \), so we set the exponent equal to 7:
\[ 40 - 3r = 7 \]

### Step 4: Solve for \( r \)
Rearranging the equation gives:
\[ 3r = 40 - 7 \\ 3r = 33 \\ r = 11 \]

### Step 5: Substitute \( r \) Back into the General Term
Now that we have \( r = 11 \), we substitute this value back into the general term to find the coefficient:
\[ T_{12} = \binom{20}{11} 2^{20-11} (-1)^{11} x^{40 - 3(11)} \]
This simplifies to:
\[ T_{12} = \binom{20}{11} 2^9 (-1)^{11} x^7 \]

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

### Step 7: Final Calculation
Now we can calculate \( \binom{20}{11} \) and \( 2^9 \):
\[ \binom{20}{11} = \frac{20!}{11!(20-11)!} = \frac{20!}{11!9!} \]
Calculating \( 2^9 = 512 \).

Thus, the final coefficient is:
\[ -\binom{20}{11} \cdot 512 \]