Find the 10th term in the binomial expansion of `(2x^2+1/x)^(12)dot`

Explanation

To find the 10th term in the binomial expansion of \( (2x^2 + \frac{1}{x})^{12} \), we can follow these steps:

### Step 1: Identify the general term in the binomial expansion
The general term \( T_{r+1} \) in the binomial expansion of \( (p + q)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} p^{n-r} q^r \]
In our case, \( p = 2x^2 \), \( q = \frac{1}{x} \), and \( n = 12 \).

### Step 2: Write the expression for the 10th term
To find the 10th term, we need to set \( r = 9 \) (since \( T_{10} = T_{r+1} \)):
\[ T_{10} = \binom{12}{9} (2x^2)^{12-9} \left(\frac{1}{x}\right)^9 \]

### Step 3: Simplify the expression
Substituting the values:
\[ T_{10} = \binom{12}{9} (2x^2)^3 \left(\frac{1}{x}\right)^9 \]

### Step 4: Calculate \( \binom{12}{9} \)
Using the property of combinations:
\[ \binom{12}{9} = \binom{12}{3} = \frac{12!}{3!(12-3)!} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220 \]

### Step 5: Calculate \( (2x^2)^3 \)
Calculating \( (2x^2)^3 \):
\[ (2x^2)^3 = 2^3 (x^2)^3 = 8x^6 \]

### Step 6: Calculate \( \left(\frac{1}{x}\right)^9 \)
Calculating \( \left(\frac{1}{x}\right)^9 \):
\[ \left(\frac{1}{x}\right)^9 = x^{-9} \]

### Step 7: Combine the results
Now we combine everything:
\[ T_{10} = 220 \cdot 8x^6 \cdot x^{-9} = 220 \cdot 8 \cdot x^{6 - 9} = 220 \cdot 8 \cdot x^{-3} \]

### Step 8: Calculate the final coefficient
Calculating \( 220 \cdot 8 \):
\[ 220 \cdot 8 = 1760 \]

### Final Result
Thus, the 10th term in the binomial expansion is:
\[ T_{10} = 1760 x^{-3} \]