The coefficient of `x^2` in the expansion of `(3x - (1)/(x) )^6` is

Options

A. 405
B. 1215
C. 2430
D. 3645

Answer

To find the coefficient of \( x^2 \) in the expansion of \( (3x - \frac{1}{x})^6 \), we can use the Binomial Theorem. The Binomial Theorem states that:

\[ (a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r \]

In our case, we have \( a = 3x \) and \( b = -\frac{1}{x} \), with \( n = 6 \).

### Step 1: Identify the general term
The general term \( T_{r+1} \) in the expansion can be expressed as:

\[ T_{r+1} = \binom{6}{r} (3x)^{6-r} \left(-\frac{1}{x}\right)^r \]

### Step 2: Simplify the general term
Now, simplifying \( T_{r+1} \):

\[ T_{r+1} = \binom{6}{r} (3^{6-r} x^{6-r}) \left(-1^r \cdot \frac{1}{x^r}\right) \]
\[ = \binom{6}{r} 3^{6-r} (-1)^r x^{6-r} \cdot x^{-r} \]
\[ = \binom{6}{r} 3^{6-r} (-1)^r x^{6-2r} \]

### Step 3: Find the term where the power of \( x \) is 2
We need the power of \( x \) to be 2:

\[ 6 - 2r = 2 \]

### Step 4: Solve for \( r \)
Solving for \( r \):

\[ 6 - 2r = 2 \implies 2r = 4 \implies r = 2 \]

### Step 5: Substitute \( r \) back into the general term
Now, substituting \( r = 2 \) back into the general term:

\[ T_{3} = \binom{6}{2} 3^{6-2} (-1)^2 x^{6-2 \cdot 2} \]
\[ = \binom{6}{2} 3^4 x^2 \]

### Step 6: Calculate the coefficient
Now, calculate \( \binom{6}{2} \) and \( 3^4 \):

\[ \binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15 \]
\[ 3^4 = 81 \]

Thus, the coefficient of \( x^2 \) is:

\[ \text{Coefficient} = 15 \times 81 = 1215 \]

### Final Answer
The coefficient of \( x^2 \) in the expansion of \( (3x - \frac{1}{x})^6 \) is **1215**.

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