If the constant term in the expansion of (1+2 | vedclass

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(B) The given expression is $(1+2x-3x^3)(\frac{3}{2}x^2-\frac{1}{3x})^9$.The general term of the expansion $(\frac{3}{2}x^2-\frac{1}{3x})^9$ is given

Vedclass · Accepted Answer

$54$

Vedclass · Answer

(B) The given expression is $(1+2x-3x^3)(\frac{3}{2}x^2-\frac{1}{3x})^9$.The general term of the expansion $(\frac{3}{2}x^2-\frac{1}{3x})^9$ is given by $T_{r+1} = {}^9C_r (\frac{3}{2}x^2)^{9-r} (-\frac{1}{3x})^r$.$T_{r+1} = {}^9C_r (\frac{3}{2})^{9-r} (-\frac{1}{3})^r x^{18-2r-r} = {}^9C_r \frac{3^{9-2r}}{2^{9-r}} (-1)^r x^{18-3r}$.To find the constant term in the product $(1+2x-3x^3)(\dots)$,we need the coefficients of $x^0$,$x^{-1}$,and $x^{-3}$ from the expansion of $(\frac{3}{2}x^2-\frac{1}{3x})^9$.$1$. For $x^0$: $18-3r = 0 \implies r=6$. Coefficient is ${}^9C_6 (\frac{3}{2})^3 (-\frac{1}{3})^6 = 84 \cdot \frac{27}{8} \cdot \frac{1}{729} = \frac{84 \cdot 27}{8 \cdot 729} = \frac{84}{8 \cdot 27} = \frac{21}{2 \cdot 27} = \frac{7}{18}$.$2$. For $x^{-1}$: $18-3r = -1$ (No integer solution for $r$).$3$. For $x^{-3}$: $18-3r = -3 \implies 3r=21 \implies r=7$. Coefficient is ${}^9C_7 (\frac{3}{2})^2 (-\frac{1}{3})^7 = 36 \cdot \frac{9}{4} \cdot (-\frac{1}{2187}) = 81 \cdot (-\frac{1}{2187}) = -\frac{1}{27}$.The constant term $p$ is obtained by: $(1 \cdot 	ext{coeff of } x^0) + (-3x^3 \cdot 	ext{coeff of } x^{-3}) = 1(\frac{7}{18}) - 3(-\frac{1}{27}) = \frac{7}{18} + \frac{1}{9} = \frac{7+2}{18} = \frac{9}{18} = \frac{1}{2}$.Therefore,$108p = 108 \cdot \frac{1}{2} = 54$.

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

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