Find the term in the expansion of `(2x^(2)-(3)/(x))^(11)` Which contains `x^(6)`

Explanation

To find the term in the expansion of \( (2x^2 - \frac{3}{x})^{11} \) that contains \( x^6 \), we will follow these steps:

### Step 1: Identify the General Term
In the binomial expansion of \( (a + b)^n \), the general term \( T_{r+1} \) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
For our expression, let \( a = 2x^2 \) and \( b = -\frac{3}{x} \), and \( n = 11 \).

### Step 2: Write the General Term for Our Expression
The general term in the expansion of \( (2x^2 - \frac{3}{x})^{11} \) is:
\[ T_{r+1} = \binom{11}{r} (2x^2)^{11-r} \left(-\frac{3}{x}\right)^r \]

### Step 3: Simplify the General Term
Now, simplify the general term:
\[ T_{r+1} = \binom{11}{r} (2^{11-r} (x^2)^{11-r}) \left(-3^r \frac{1}{x^r}\right) \]
This simplifies to:
\[ T_{r+1} = \binom{11}{r} (-3)^r 2^{11-r} x^{22 - r} \]

### Step 4: Set the Power of \( x \) to 6
We want the term where the power of \( x \) is 6:
\[ 22 - r = 6 \]

### Step 5: Solve for \( r \)
Rearranging gives:
\[ r = 22 - 6 = 16 \]

### Step 6: Check if \( r \) is Valid
Since \( r \) must be a non-negative integer and \( r = 16 \) is greater than \( n = 11 \), this means that there is no valid term for \( r = 16 \).

### Conclusion
Thus, the term in the expansion of \( (2x^2 - \frac{3}{x})^{11} \) that contains \( x^6 \) does not exist.

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