If the term independent of x in the expansion of `(2+5x+ax^(3)) ((3)/(2)x^(2)-(1)/(3x))^(9)` is 1, then a is equal to

Options

A. 3
B. 4
C. 5
D. `-6`

Answer

To find the value of \( a \) such that the term independent of \( x \) in the expansion of \( (2 + 5x + ax^3) \left( \frac{3}{2}x^2 - \frac{1}{3x} \right)^9 \) is equal to 1, we will follow these steps:

### Step 1: Expand the second factor
We start with the second factor \( \left( \frac{3}{2}x^2 - \frac{1}{3x} \right)^9 \). We can use the binomial theorem to expand this expression:

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

Here, \( a = \frac{3}{2}x^2 \) and \( b = -\frac{1}{3x} \), and \( n = 9 \).

### Step 2: Write the general term
The general term in the expansion will be:

\[ T_r = \binom{9}{r} \left( \frac{3}{2}x^2 \right)^{9-r} \left( -\frac{1}{3x} \right)^r \]

This simplifies to:

\[ T_r = \binom{9}{r} \left( \frac{3}{2} \right)^{9-r} \left( -\frac{1}{3} \right)^r x^{2(9-r) - r} \]

\[ = \binom{9}{r} \left( \frac{3}{2} \right)^{9-r} \left( -\frac{1}{3} \right)^r x^{18 - 3r} \]

### Step 3: Combine with the first factor
Now we need to combine this with the first factor \( (2 + 5x + ax^3) \). We will consider the contributions from each term in \( (2 + 5x + ax^3) \).

1. **From the term \( 2 \)**:
- The term from \( T_r \) must have \( x^{0} \), so we need \( 18 - 3r = 0 \) which gives \( r = 6 \).

2. **From the term \( 5x \)**:
- The term from \( T_r \) must have \( x^{-1} \), so we need \( 18 - 3r = -1 \) which gives \( r = 7 \).

3. **From the term \( ax^3 \)**:
- The term from \( T_r \) must have \( x^{-3} \), so we need \( 18 - 3r = -3 \) which gives \( r = 7 \).

### Step 4: Calculate the coefficients
Now we calculate the coefficients for \( r = 6 \) and \( r = 7 \).

- For \( r = 6 \):
\[ T_6 = \binom{9}{6} \left( \frac{3}{2} \right)^{3} \left( -\frac{1}{3} \right)^{6} \]
\[ = \binom{9}{6} \cdot \frac{27}{8} \cdot \left( -\frac{1}{729} \right) \]
\[ = \binom{9}{6} \cdot \frac{27 \cdot (-1)}{8 \cdot 729} \]

- For \( r = 7 \):
\[ T_7 = \binom{9}{7} \left( \frac{3}{2} \right)^{2} \left( -\frac{1}{3} \right)^{7} \]
\[ = \binom{9}{7} \cdot \frac{9}{4} \cdot \left( -\frac{1}{2187} \right) \]
\[ = \binom{9}{7} \cdot \frac{9 \cdot (-1)}{4 \cdot 2187} \]

### Step 5: Set up the equation
The total contribution of the independent term is:

\[ 2 \cdot T_6 + 5 \cdot T_7 + a \cdot T_7 = 1 \]

### Step 6: Solve for \( a \)
We can now substitute the values of \( T_6 \) and \( T_7 \) into the equation and solve for \( a \).