Question

If the constant term in the expansion of $(1 + 2x - 3x^3) \left(\frac{3}{2}x^2 - \frac{1}{3x}\right)^9$ is $p$, then $108p$ is equal to _________.

Answer 1

Correct Answer: 54

Given expression:

$$\left(1 + 2x - 3x^3\right)\left(\frac{3}{2}x^2 - \frac{1}{3x}\right)^9$$

General term:

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

Simplifying:

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

To find the constant term, we set the exponent of $x$ to zero:

$$18 - 3r = 0 \implies r = 6$$

Substituting $r = 6$ into the expression:

$$T_6 = \binom{9}{6} \left(\frac{3}{2}\right)^3 \left(-\frac{1}{3}\right)^6$$

Calculating each term:

$$\binom{9}{6} = 84, \quad \left(\frac{3}{2}\right)^3 = \frac{27}{8}, \quad \left(-\frac{1}{3}\right)^6 = \frac{1}{729}$$ $$T_6 = 84 \times \frac{27}{8} \times \frac{1}{729} = \frac{7}{18}$$

Next, substituting $r = 7$ to find the coefficient of $x^{-3}$:

$$T_7 = \binom{9}{7} \left(\frac{3}{2}\right)^2 \left(-\frac{1}{3}\right)^7$$

Calculating:

$$\binom{9}{7} = 36, \quad \left(\frac{3}{2}\right)^2 = \frac{9}{4}, \quad \left(-\frac{1}{3}\right)^7 = -\frac{1}{2187}$$ $$T_7 = 36 \times \frac{9}{4} \times -\frac{1}{2187} = -\frac{1}{27}$$

Combining the terms:

$$\left(1 + 2x - 3x^3\right)\left(\frac{7}{18} + \frac{-1}{27}x^3\right)$$

Simplifying:

$$\text{Constant term} = \frac{7}{18}$$

Given that $p$ is the constant term, we have $p = \frac{7}{18}$. Calculating $108p$:

$$108p = 108 \times \frac{7}{18} = 54$$