The coefficient of the term independent of x | vedclass

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(C) The $(r+1)$-th term in the expansion of $\left( \frac{3}{2}x^2 - \frac{1}{3x} \right)^9$ is given by$T_{r+1} = \binom{9}{r} \left( \frac{3}{2}x^2

Vedclass · Accepted Answer

$17/54$

Vedclass · Answer

(C) The $(r+1)$-th term in the expansion of $\left( \frac{3}{2}x^2 - \frac{1}{3x} \right)^9$ is given by$T_{r+1} = \binom{9}{r} \left( \frac{3}{2}x^2 \right)^{9-r} \left( -\frac{1}{3x} \right)^r = \binom{9}{r} (-1)^r \frac{3^{9-2r}}{2^{9-r}} x^{18-3r}$ .....$(1)$We need the coefficient of the term independent of $x$ in $(1+x+2x^3) \left( \frac{3}{2}x^2 - \frac{1}{3x} \right)^9$.This requires finding the coefficients of $x^0$,$x^{-1}$,and $x^{-3}$ in the expansion of $\left( \frac{3}{2}x^2 - \frac{1}{3x} \right)^9$.For $x^0$,$18-3r=0 \implies r=6$.For $x^{-1}$,$18-3r=-1 \implies 3r=19$ (no integer solution).For $x^{-3}$,$18-3r=-3 \implies 3r=21 \implies r=7$.The coefficient of the term independent of $x$ is:$1 \cdot \binom{9}{6} (-1)^6 \frac{3^{9-12}}{2^{9-6}} + 2 \cdot \binom{9}{7} (-1)^7 \frac{3^{9-14}}{2^{9-7}}$$= \binom{9}{3} \frac{3^{-3}}{2^3} - 2 \binom{9}{2} \frac{3^{-5}}{2^2} = \frac{84}{8 \cdot 27} - \frac{2 \cdot 36}{4 \cdot 243} = \frac{84}{216} - \frac{72}{972} = \frac{7}{18} - \frac{2}{27} = \frac{21-4}{54} = \frac{17}{54}$.

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

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