Prove that the coefficient of `x^3` in the expansion of `(1+x+2x^2) (2x^2 - 1/(3x))^9` is `-224/27.`

Answer

To find the coefficient of \( x^3 \) in the expansion of \( (1 + x + 2x^2)(2x^2 - \frac{1}{3x})^9 \), we will break down the problem into manageable steps.

### Step 1: Expand \( (2x^2 - \frac{1}{3x})^9 \)

Using the Binomial Theorem, the general term in the expansion of \( (a + b)^n \) is given by:

\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]

In our case, \( a = 2x^2 \), \( b = -\frac{1}{3x} \), and \( n = 9 \). Thus, the general term \( T_{r+1} \) can be expressed as:

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

This simplifies to:

\[ T_{r+1} = \binom{9}{r} 2^{9-r} (-1)^r \frac{x^{2(9-r)}}{3^r x^r} \]

Combining the powers of \( x \):

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

### Step 2: Find the coefficient of \( x^3 \)

We need \( 18 - 3r = 3 \) to find the relevant term for \( x^3 \):

\[ 18 - 3r = 3 \implies 3r = 15 \implies r = 5 \]

Now substituting \( r = 5 \) into the general term:

\[ T_{6} = \binom{9}{5} 2^{4} (-1)^5 \frac{1}{3^5} x^3 \]

Calculating \( \binom{9}{5} \):

\[ \binom{9}{5} = \frac{9!}{5!(9-5)!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126 \]

Thus,

\[ T_{6} = 126 \cdot 16 \cdot (-1) \cdot \frac{1}{243} x^3 = -\frac{2016}{243} x^3 \]

### Step 3: Coefficient from \( (1 + x + 2x^2) \)

Now we need to consider the contributions from \( (1 + x + 2x^2) \):

1. **From \( 1 \)**: The coefficient of \( x^3 \) from \( T_6 \) is \(-\frac{2016}{243}\).
2. **From \( x \)**: We need the coefficient of \( x^2 \) from \( T_5 \):

\[ 18 - 3r = 2 \implies 3r = 16 \implies r = \frac{16}{3} \quad (\text{not possible}) \]

3. **From \( 2x^2 \)**: We need the coefficient of \( x^1 \) from \( T_4 \):

\[ 18 - 3r = 1 \implies 3r = 17 \implies r = \frac{17}{3} \quad (\text{not possible}) \]

Thus, the only contribution to the coefficient of \( x^3 \) comes from the term \( 1 \).

### Step 4: Final Calculation

The coefficient of \( x^3 \) in the expansion of \( (1 + x + 2x^2)(2x^2 - \frac{1}{3x})^9 \) is:

\[ -\frac{2016}{243} = -\frac{224}{27} \]

### Conclusion

Thus, we have proved that the coefficient of \( x^3 \) in the expansion is \( -\frac{224}{27} \).

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