The fourth term in the binomial expansion of `(x^2 - (1)/(x^3) )^n` is independent of x , when n is equal to

Options

A. 2
B. 3
C. 4
D. none of these

Answer

To find the value of \( n \) such that the fourth term in the binomial expansion of \( (x^2 - \frac{1}{x^3})^n \) is independent of \( x \), we can follow these steps:

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

### Step 2: Write the Fourth Term
To find the fourth term, we need to set \( r = 3 \) (since the first term corresponds to \( r = 0 \)):
\[ T_4 = \binom{n}{3} (x^2)^{n-3} \left(-\frac{1}{x^3}\right)^3 \]

### Step 3: Simplify the Fourth Term
Now, we simplify \( T_4 \):
\[ T_4 = \binom{n}{3} (x^2)^{n-3} \left(-\frac{1}{x^3}\right)^3 = \binom{n}{3} (x^2)^{n-3} \left(-\frac{1}{x^9}\right) \]
This can be rewritten as:
\[ T_4 = \binom{n}{3} (-1) (x^{2(n-3)}) (x^{-9}) = -\binom{n}{3} x^{2n - 6 - 9} = -\binom{n}{3} x^{2n - 15} \]

### Step 4: Set the Exponent of \( x \) to Zero
For the term to be independent of \( x \), the exponent must be zero:
\[ 2n - 15 = 0 \]

### Step 5: Solve for \( n \)
Now, we solve for \( n \):
\[ 2n = 15 \implies n = \frac{15}{2} \]

### Final Answer
Thus, the value of \( n \) such that the fourth term is independent of \( x \) is:
\[ \boxed{\frac{15}{2}} \]