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, which states that:

\[ (a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r \]

In our case, let \( a = 2x^2 \) and \( b = -\frac{1}{x} \), and \( n = 20 \).

### Step 1: Set up the binomial expansion
Using the binomial theorem, we can express the expansion as:

\[ (2x^2 - \frac{1}{x})^{20} = \sum_{r=0}^{20} \binom{20}{r} (2x^2)^{20-r} \left(-\frac{1}{x}\right)^r \]

### Step 2: Simplify the terms
Now, we need to simplify the general term in the expansion:

\[ T_r = \binom{20}{r} (2x^2)^{20-r} \left(-\frac{1}{x}\right)^r \]

This becomes:

\[ T_r = \binom{20}{r} (2^{20-r} (x^2)^{20-r}) \left(-1\right)^r (x^{-r}) \]

Which simplifies to:

\[ T_r = \binom{20}{r} (-1)^r 2^{20-r} x^{2(20-r) - r} \]

### Step 3: Determine the power of \( x \)
We need the power of \( x \) to equal 7:

\[ 2(20 - r) - r = 7 \]

Expanding this gives:

\[ 40 - 2r - r = 7 \]

This simplifies to:

\[ 40 - 3r = 7 \]

### Step 4: Solve for \( r \)
Rearranging gives:

\[ 3r = 40 - 7 = 33 \implies r = \frac{33}{3} = 11 \]

### Step 5: Find the coefficient
Now that we have \( r = 11 \), we can substitute \( r \) back into the expression for \( T_r \):

\[ T_{11} = \binom{20}{11} (-1)^{11} 2^{20-11} x^{7} \]

Calculating the coefficient:

\[ \text{Coefficient} = \binom{20}{11} (-1)^{11} 2^{9} \]

Since \( (-1)^{11} = -1 \), we have:

\[ \text{Coefficient} = -\binom{20}{11} \cdot 2^9 \]

### Step 6: Calculate \( \binom{20}{11} \) and \( 2^9 \)
Calculating \( \binom{20}{11} \):

\[ \binom{20}{11} = \frac{20!}{11! \cdot 9!} \]

And \( 2^9 = 512 \).

### Final Step: Combine the results
Thus, the coefficient of \( x^7 \) is:

\[ -\binom{20}{11} \cdot 512 \]