Question

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

• A. $-\frac{224}{9}$
• B. $-\frac{112}{9}$
• C. $-\frac{224}{27}$
• D. $-\frac{112}{27}$

Answer 1

Answer: (C)

The coefficient of $x^3$ in the expansion of $(1+x+2x^2)\left(2x^2-\frac{1}{3x}\right)^9$ is $-\frac{224}{27}$.

Answer 2

The general term in the expansion of $\left(2x^2-\frac{1}{3x}\right)^9$ is $T_{r+1} = \binom{9}{r} (2x^2)^{9-r} \left(-\frac{1}{3x}\right)^r = \binom{9}{r} 2^{9-r} \left(-\frac{1}{3}\right)^r x^{18-3r}$. To find the coefficient of $x^3$ in the product $(1+x+2x^2)\left(2x^2-\frac{1}{3x}\right)^9$, we consider the terms in $(1+x+2x^2)$:

1. From $1$: we need $x^3$ from the binomial expansion. $18-3r = 3 \implies r=5$. Coefficient is $\binom{9}{5} 2^4 \left(-\frac{1}{3}\right)^5 = 126 \times 16 \times (-\frac{1}{243}) = -\frac{2016}{243} = -\frac{224}{27}$.
2. From $x$: we need $x^2$ from the binomial expansion. $18-3r = 2 \implies 3r=16$, $r$ is not an integer.
3. From $2x^2$: we need $x^1$ from the binomial expansion. $18-3r = 1 \implies 3r=17$, $r$ is not an integer. The total coefficient of $x^3$ is $-\frac{224}{27}$.