Find the term independent of `x` in the expansion of `(3x^2-1/(2x^3))^(10)` .

Options

A. 5th term
B. 6th Term
C. 7th term
D. 4th term

Explanation

To find the term independent of \( x \) in the expansion of \( (3x^2 - \frac{1}{2x^3})^{10} \), we can follow these steps:

### Step 1: Identify the General Term
In the binomial expansion of \( (a + b)^n \), the general term \( T_r \) is given by:
\[ T_r = \binom{n}{r} a^{n-r} b^r \]
For our expression, \( a = 3x^2 \) and \( b = -\frac{1}{2x^3} \), and \( n = 10 \).

### Step 2: Write the General Term for the Given Expression
Substituting the values into the general term formula, we get:
\[ T_r = \binom{10}{r} (3x^2)^{10-r} \left(-\frac{1}{2x^3}\right)^r \]

### Step 3: Simplify the General Term
Now, simplifying \( T_r \):
\[ T_r = \binom{10}{r} (3^{10-r} (x^2)^{10-r}) \left(-\frac{1}{2^r (x^3)^r}\right) \]
\[ = \binom{10}{r} (-1)^r \frac{3^{10-r}}{2^r} x^{2(10-r) - 3r} \]
\[ = \binom{10}{r} (-1)^r \frac{3^{10-r}}{2^r} x^{20 - 2r - 3r} \]
\[ = \binom{10}{r} (-1)^r \frac{3^{10-r}}{2^r} x^{20 - 5r} \]

### Step 4: Find the Term Independent of \( x \)
For the term to be independent of \( x \), the exponent of \( x \) must be zero:
\[ 20 - 5r = 0 \]
Solving for \( r \):
\[ 5r = 20 \implies r = 4 \]

### Step 5: Substitute \( r \) Back into the General Term
Now, we substitute \( r = 4 \) back into the general term to find the term independent of \( x \):
\[ T_4 = \binom{10}{4} (-1)^4 \frac{3^{10-4}}{2^4} x^{20 - 5 \cdot 4} \]
\[ = \binom{10}{4} \frac{3^6}{2^4} \]

### Step 6: Calculate \( \binom{10}{4} \), \( 3^6 \), and \( 2^4 \)
Calculating each component:
\[ \binom{10}{4} = \frac{10!}{4!(10-4)!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210 \]
\[ 3^6 = 729 \]
\[ 2^4 = 16 \]

### Step 7: Final Calculation
Now substituting these values:
\[ T_4 = 210 \cdot \frac{729}{16} \]
Calculating this gives:
\[ T_4 = \frac{210 \times 729}{16} = \frac{1536090}{16} = 96068.125 \]

Thus, the term independent of \( x \) is:
\[ \frac{1536090}{16} \]