Question

Find the term independent of 'x' in the expansion of the expression, $(1 + x + 2 x^3)\left(\frac{3}{2}x^2 - \frac{1}{3x}\right)^9.$

Answer 1

17/54

Answer 2

Let the given expression be $E$. $E = (1 + x + 2 x^3)\left(\frac{3}{2}x^2 - \frac{1}{3x}\right)^9.$ We need to find the term independent of 'x' in the expansion of $E$. The expression is a product of two factors: $(1 + x + 2 x^3)$ and $\left(\frac{3}{2}x^2 - \frac{1}{3x}\right)^9$. Let's find the general term in the expansion of the second factor using the binomial theorem. The general term in the expansion of $(a+b)^n$ is $T_{r+1} = \binom{n}{r} a^{n-r} b^r$. Here, $a = \frac{3}{2}x^2$, $b = -\frac{1}{3x}$, and $n=9$. The general term in the expansion of $\left(\frac{3}{2}x^2 - \frac{1}{3x}\right)^9$ is: $T_{r+1} = \binom{9}{r} \left(\frac{3}{2}x^2\right)^{9-r} \left(-\frac{1}{3x}\right)^r$ $T_{r+1} = \binom{9}{r} \left(\frac{3}{2}\right)^{9-r} (x^2)^{9-r} \left(-\frac{1}{3}\right)^r \left(\frac{1}{x}\right)^r$ $T_{r+1} = \binom{9}{r} \left(\frac{3}{2}\right)^{9-r} x^{18-2r} \left(-\frac{1}{3}\right)^r x^{-r}$ $T_{r+1} = \binom{9}{r} \left(\frac{3}{2}\right)^{9-r} \left(-\frac{1}{3}\right)^r x^{18-2r-r}$ $T_{r+1} = \binom{9}{r} \left(\frac{3}{2}\right)^{9-r} \left(-\frac{1}{3}\right)^r x^{18-3r}, \quad \text{for } r = 0, 1, \ldots, 9.$

The expression $E$ is $(1 + x + 2 x^3) \times \sum_{r=0}^9 T_{r+1}$. To find the term independent of 'x', we need to consider the product of terms from the first factor and the second factor such that the powers of 'x' sum to 0.

The first factor is $1 + x + 2x^3$. The terms are $1x^0$, $1x^1$, and $2x^3$.

Case 1: Term from the first factor is $1$ (which is $1x^0$). We need the term independent of 'x' from $1 \times \left(\frac{3}{2}x^2 - \frac{1}{3x}\right)^9$. This means we need the term with $x^0$ in the expansion of $\left(\frac{3}{2}x^2 - \frac{1}{3x}\right)^9$. The power of 'x' in the general term $T_{r+1}$ is $18-3r$. For the term to be independent of 'x', $18-3r = 0$, which gives $3r=18$, so $r=6$. Since $r=6$ is an integer between 0 and 9, this term exists. The term is $T_{6+1} = T_7$. The coefficient of $x^0$ in this case is $1 \times \binom{9}{6} \left(\frac{3}{2}\right)^{9-6} \left(-\frac{1}{3}\right)^6$. Coefficient = $\binom{9}{6} \left(\frac{3}{2}\right)^3 \left(-\frac{1}{3}\right)^6 = \binom{9}{3} \left(\frac{27}{8}\right) \left(\frac{1}{729}\right)$. $\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$. Coefficient = $84 \times \frac{27}{8} \times \frac{1}{729} = \frac{84 \times 27}{8 \times 729}$. Since $729 = 27 \times 27$, we have $\frac{84 \times 27}{8 \times 27 \times 27} = \frac{84}{8 \times 27}$. $\frac{84}{8 \times 27} = \frac{21 \times 4}{2 \times 4 \times 27} = \frac{21}{2 \times 27} = \frac{7 \times 3}{2 \times 9 \times 3} = \frac{7}{18}$.

Case 2: Term from the first factor is $x$ (which is $1x^1$). We need the term independent of 'x' from $x \times \left(\frac{3}{2}x^2 - \frac{1}{3x}\right)^9$. This means we need the term with $x^{-1}$ from the expansion of $\left(\frac{3}{2}x^2 - \frac{1}{3x}\right)^9$. The power of 'x' in the general term $T_{r+1}$ is $18-3r$. We need $18-3r = -1$, which gives $3r=19$, so $r=19/3$. Since $r=19/3$ is not an integer, there is no such term in the expansion. The contribution to the term independent of 'x' is 0.

Case 3: Term from the first factor is $2x^3$ (which is $2x^3$). We need the term independent of 'x' from $2x^3 \times \left(\frac{3}{2}x^2 - \frac{1}{3x}\right)^9$. This means we need the term with $x^{-3}$ from the expansion of $\left(\frac{3}{2}x^2 - \frac{1}{3x}\right)^9$. The power of 'x' in the general term $T_{r+1}$ is $18-3r$. We need $18-3r = -3$, which gives $3r=21$, so $r=7$. Since $r=7$ is an integer between 0 and 9, this term exists. The term is $T_{7+1} = T_8$. The coefficient of $x^{-3}$ in the expansion of $\left(\frac{3}{2}x^2 - \frac{1}{3x}\right)^9$ is $\binom{9}{7} \left(\frac{3}{2}\right)^{9-7} \left(-\frac{1}{3}\right)^7$. The coefficient of $x^0$ in this case is $2 \times \binom{9}{7} \left(\frac{3}{2}\right)^2 \left(-\frac{1}{3}\right)^7$. Coefficient = $2 \times \binom{9}{2} \left(\frac{9}{4}\right) \left(-\frac{1}{2187}\right)$. $\binom{9}{2} = \frac{9 \times 8}{2 \times 1} = 36$. Coefficient = $2 \times 36 \times \frac{9}{4} \times \left(-\frac{1}{2187}\right) = 72 \times \frac{9}{4} \times \left(-\frac{1}{2187}\right)$. $72 \times \frac{9}{4} = 18 \times 9 = 162$. Coefficient = $162 \times \left(-\frac{1}{2187}\right) = -\frac{162}{2187}$. We simplify the fraction: $162 = 2 \times 81 = 2 \times 3^4$, and $2187 = 3^7$. Coefficient = $-\frac{2 \times 3^4}{3^7} = -\frac{2}{3^{7-4}} = -\frac{2}{3^3} = -\frac{2}{27}$.

The term independent of 'x' in the expansion of the entire expression is the sum of the coefficients from these three cases. Total term independent of 'x' = (Coefficient from Case 1) + (Coefficient from Case 2) + (Coefficient from Case 3). Total term independent of 'x' = $\frac{7}{18} + 0 + \left(-\frac{2}{27}\right) = \frac{7}{18} - \frac{2}{27}$.

To subtract the fractions, we find a common denominator, which is LCM(18, 27) = 54. $\frac{7}{18} = \frac{7 \times 3}{18 \times 3} = \frac{21}{54}$. $\frac{2}{27} = \frac{2 \times 2}{27 \times 2} = \frac{4}{54}$. Total term independent of 'x' = $\frac{21}{54} - \frac{4}{54} = \frac{21 - 4}{54} = \frac{17}{54}$.

The final answer is $\frac{17}{54}$.