The term independent of x in the expansion of `(x-3/x^(2))^(18)` is

Options

A. `.^(18)c_(6)0`
B. `.^(18)c_(6)3^(6)`
C. `.^(18)c_(12)`
D. `.^(18)c_(6)3^(12)`

Answer

To find the term independent of \( x \) in the expansion of \( \left( x - \frac{3}{x^2} \right)^{18} \), we can follow these steps:

### Step 1: Identify the general term
The general term \( T_{r+1} \) 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, \( a = x \) and \( b = -\frac{3}{x^2} \), and \( n = 18 \). Thus, the general term can be expressed as:
\[ T_{r+1} = \binom{18}{r} x^{18 - r} \left(-\frac{3}{x^2}\right)^r \]

### Step 2: Simplify the general term
Now, we simplify \( T_{r+1} \):
\[ T_{r+1} = \binom{18}{r} x^{18 - r} \cdot \left(-3\right)^r \cdot \frac{1}{x^{2r}} = \binom{18}{r} (-3)^r x^{18 - r - 2r} \]
This simplifies to:
\[ T_{r+1} = \binom{18}{r} (-3)^r x^{18 - 3r} \]

### Step 3: Find the term independent of \( x \)
For the term to be independent of \( x \), we need the exponent of \( x \) to be zero:
\[ 18 - 3r = 0 \]
Solving for \( r \):
\[ 3r = 18 \implies r = 6 \]

### Step 4: Identify the term corresponding to \( r = 6 \)
Now, we substitute \( r = 6 \) back into the general term to find the term independent of \( x \):
\[ T_{7} = \binom{18}{6} (-3)^6 \]

### Step 5: Calculate the value
Calculating \( (-3)^6 \):
\[ (-3)^6 = 729 \]
Now we need to calculate \( \binom{18}{6} \):
\[ \binom{18}{6} = \frac{18!}{6!(18-6)!} = \frac{18!}{6! \cdot 12!} \]
Calculating this gives:
\[ \binom{18}{6} = 18564 \]
Now, we multiply:
\[ T_{7} = 18564 \cdot 729 \]

### Step 6: Final calculation
Calculating \( 18564 \cdot 729 \):
\[ T_{7} = 13520736 \]

Thus, the term independent of \( x \) in the expansion of \( \left( x - \frac{3}{x^2} \right)^{18} \) is \( 13520736 \).