The coefficient of `x^(7)` in the expansion of `( ax^(2) + ( 1)/( bx))^(8)` is

Options

A. `""_(8)C_(3) (a^(3))/(b^(5))`
B. `""_(8)C_(3) (a^(5))/(b^(3))`
C. `""_(8)C_(4) (a^(5))/(b^(3))`
D. None of these

Answer

To find the coefficient of \( x^7 \) in the expansion of \( (ax^2 + \frac{1}{bx})^8 \), 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, \( a = ax^2 \), \( b = \frac{1}{bx} \), and \( n = 8 \).

### Step 1: Identify the general term
The general term \( T_r \) in the expansion is given by:

\[ T_r = \binom{8}{r} (ax^2)^{8-r} \left(\frac{1}{bx}\right)^r \]

### Step 2: Simplify the general term
Now, we simplify \( T_r \):

\[ T_r = \binom{8}{r} (a^{8-r} (x^2)^{8-r}) \left(\frac{1}{b^r x^r}\right) \]

This expands to:

\[ T_r = \binom{8}{r} a^{8-r} \frac{x^{2(8-r)}}{b^r x^r} = \binom{8}{r} a^{8-r} \frac{x^{16 - 2r}}{b^r} \]

### Step 3: Find the power of \( x \)
We want the coefficient of \( x^7 \). Therefore, we need to solve for \( r \) in the equation:

\[ 16 - 2r - r = 7 \]

This simplifies to:

\[ 16 - 3r = 7 \]

### Step 4: Solve for \( r \)
Rearranging gives:

\[ 3r = 9 \implies r = 3 \]

### Step 5: Substitute \( r \) back into the general term
Now, substitute \( r = 3 \) back into the general term:

\[ T_3 = \binom{8}{3} a^{8-3} \frac{1}{b^3} x^{16 - 2 \cdot 3} \]

This simplifies to:

\[ T_3 = \binom{8}{3} a^5 \frac{1}{b^3} x^7 \]

### Step 6: Find the coefficient
The coefficient of \( x^7 \) is:

\[ \binom{8}{3} a^5 \frac{1}{b^3} \]

### Step 7: Calculate \( \binom{8}{3} \)
Calculating \( \binom{8}{3} \):

\[ \binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56 \]

### Final Coefficient
Thus, the coefficient of \( x^7 \) is:

\[ \frac{56 a^5}{b^3} \]

### Summary
The coefficient of \( x^7 \) in the expansion of \( (ax^2 + \frac{1}{bx})^8 \) is:

\[ \frac{56 a^5}{b^3} \]

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