Find the coefficient of `x^9` and `x^10` in ` (2x^2 - 1/x)^20`

Answer

To find the coefficients of \(x^9\) and \(x^{10}\) in the expression \((2x^2 - \frac{1}{x})^{20}\), we will use the Binomial Theorem.

### Step 1: Write the general term of the binomial expansion
The general term \(T_{r+1}\) in the 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 \(n = 20\). Thus, the general term becomes:
\[ T_{r+1} = \binom{20}{r} (2x^2)^{20-r} \left(-\frac{1}{x}\right)^r \]

### Step 2: Simplify the general term
Now, we simplify \(T_{r+1}\):
\[ T_{r+1} = \binom{20}{r} (2^{20-r} (x^2)^{20-r}) \left(-\frac{1}{x}\right)^r \]
This can be rewritten as:
\[ T_{r+1} = \binom{20}{r} 2^{20-r} (-1)^r x^{2(20-r) - r} \]
\[ = \binom{20}{r} 2^{20-r} (-1)^r x^{40 - 3r} \]

### Step 3: Find the coefficients for \(x^9\) and \(x^{10}\)
We need to find the values of \(r\) such that \(40 - 3r = 9\) and \(40 - 3r = 10\).

#### For \(x^9\):
Set up the equation:
\[ 40 - 3r = 9 \]
\[ 3r = 40 - 9 = 31 \quad \Rightarrow \quad r = \frac{31}{3} \quad \text{(not an integer)} \]
Thus, there is no coefficient for \(x^9\).

#### For \(x^{10}\):
Set up the equation:
\[ 40 - 3r = 10 \]
\[ 3r = 40 - 10 = 30 \quad \Rightarrow \quad r = 10 \]

### Step 4: Calculate the coefficient for \(x^{10}\)
Substituting \(r = 10\) into the general term:
\[ T_{11} = \binom{20}{10} 2^{20-10} (-1)^{10} x^{10} \]
\[ = \binom{20}{10} 2^{10} x^{10} \]
The coefficient of \(x^{10}\) is:
\[ \binom{20}{10} \cdot 2^{10} \]

### Final Answer
- The coefficient of \(x^9\) is **0**.
- The coefficient of \(x^{10}\) is \(\binom{20}{10} \cdot 2^{10}\).