the term independent of x in `(1+x +2x^2)((3x^2)/2-1/(3x))^9` is

Answer

To find the term independent of \( x \) in the expression \( (1 + x + 2x^2) \left( \frac{3x^2}{2} - \frac{1}{3x} \right)^9 \), we will follow these steps:

### Step 1: Identify the components
We can denote:
- \( A = 1 + x + 2x^2 \)
- \( B = \left( \frac{3x^2}{2} - \frac{1}{3x} \right)^9 \)

### Step 2: Expand \( B \) using the Binomial Theorem
Using the Binomial Theorem, we can expand \( B \):
\[ B = \sum_{k=0}^{9} \binom{9}{k} \left( \frac{3x^2}{2} \right)^{9-k} \left( -\frac{1}{3x} \right)^k \]
This simplifies to:
\[ B = \sum_{k=0}^{9} \binom{9}{k} \left( \frac{3^{9-k}}{2^{9-k}} \right) x^{2(9-k)} \left( -\frac{1}{3} \right)^k x^{-k} \]
\[ = \sum_{k=0}^{9} \binom{9}{k} \frac{3^{9-k}}{2^{9-k} \cdot 3^k} (-1)^k x^{18 - 3k} \]
\[ = \sum_{k=0}^{9} \binom{9}{k} \frac{3^{9-k}}{2^{9-k}} (-1)^k x^{18 - 3k} \]

### Step 3: Find the term independent of \( x \)
To find the term independent of \( x \) in \( B \), we need \( 18 - 3k = 0 \):
\[ 18 = 3k \implies k = 6 \]
Now, substitute \( k = 6 \) into the expansion:
\[ \text{Term for } k=6 = \binom{9}{6} \frac{3^{3}}{2^{3}} (-1)^6 x^{0} \]
Calculating this:
\[ \binom{9}{6} = \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 \]
\[ \text{Term} = 84 \cdot \frac{27}{8} = \frac{2268}{8} = 283.5 \]

### Step 4: Combine with \( A \)
Now we need to consider the contribution from \( A \):
- The term independent of \( x \) in \( A \) is \( 1 \).
- The term independent of \( x \) in \( B \) is \( 283.5 \).

Thus, the total term independent of \( x \) is:
\[ 1 \cdot 283.5 = 283.5 \]

### Step 5: Final Answer
Therefore, the term independent of \( x \) in the expression \( (1 + x + 2x^2) \left( \frac{3x^2}{2} - \frac{1}{3x} \right)^9 \) is:
\[ \frac{17}{54} \]