The coefficient of `x^(-7)` in the expansion of `(ax-(1)/(bx^(2)))^(11)` will be

Options

A. `(461 a^(6))/(b^(3))`
B. `(462 a^(5))/(b^(6))`
C. `(462a^(5))/(b^(6))`
D. None of these

Answer

To find the coefficient of \( x^{-7} \) in the expansion of \( (ax - \frac{1}{bx^2})^{11} \), we can use the Binomial Theorem. The Binomial Theorem states that:

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

In our case, we can identify \( a = ax \) and \( b = -\frac{1}{bx^2} \), and \( n = 11 \).

### Step 1: Write the general term of the expansion

The general term \( T_k \) in the expansion can be expressed as:

\[ T_k = \binom{11}{k} (ax)^{11-k} \left(-\frac{1}{bx^2}\right)^k \]

### Step 2: Simplify the general term

Now, let's simplify \( T_k \):

\[ T_k = \binom{11}{k} (ax)^{11-k} \left(-\frac{1}{bx^2}\right)^k = \binom{11}{k} (a^{11-k} x^{11-k}) \left(-\frac{1}{b^k x^{2k}}\right) \]

Combining the terms, we have:

\[ T_k = \binom{11}{k} (-1)^k \frac{a^{11-k}}{b^k} x^{11-k - 2k} = \binom{11}{k} (-1)^k \frac{a^{11-k}}{b^k} x^{11 - 3k} \]

### Step 3: Set the exponent of \( x \) to \(-7\)

We want the coefficient of \( x^{-7} \), so we set the exponent equal to \(-7\):

\[ 11 - 3k = -7 \]

### Step 4: Solve for \( k \)

Now, we solve for \( k \):

\[ 11 + 7 = 3k \\ 18 = 3k \\ k = 6 \]

### Step 5: Substitute \( k \) back into the general term

Now that we have \( k = 6 \), we substitute this back into the expression for \( T_k \):

\[ T_6 = \binom{11}{6} (-1)^6 \frac{a^{11-6}}{b^6} x^{11 - 3 \cdot 6} \]

This simplifies to:

\[ T_6 = \binom{11}{6} \frac{a^5}{b^6} x^{-7} \]

### Step 6: Find the coefficient

The coefficient of \( x^{-7} \) is:

\[ \binom{11}{6} \frac{a^5}{b^6} \]

### Final Answer

Thus, the coefficient of \( x^{-7} \) in the expansion of \( (ax - \frac{1}{bx^2})^{11} \) is:

\[ \frac{11!}{6!5!} \cdot \frac{a^5}{b^6} = \frac{462 a^5}{b^6} \]