In the expansion of {left( frac{3x^2}{2} - fr | vedclass

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Question

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

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$^9C_3 \cdot \frac{1}{6^3}$

Vedclass · Answer

(A) The general term in the expansion of ${\left( \frac{3x^2}{2} - \frac{1}{3x} \right)^9}$ is given by:$T_{r+1} = ^9C_r \left( \frac{3x^2}{2} \right)^{9-r} \left( -\frac{1}{3x} \right)^r$Simplifying the expression:$T_{r+1} = ^9C_r \left( \frac{3}{2} \right)^{9-r} (x^2)^{9-r} \left( -\frac{1}{3} \right)^r (x^{-1})^r$$T_{r+1} = ^9C_r \left( \frac{3}{2} \right)^{9-r} \left( -\frac{1}{3} \right)^r x^{18-2r-r}$$T_{r+1} = ^9C_r \left( \frac{3}{2} \right)^{9-r} \left( -\frac{1}{3} \right)^r x^{18-3r}$For the term independent of $x$,the power of $x$ must be $0$:$18 - 3r = 0 \implies 3r = 18 \implies r = 6$Substituting $r = 6$ into the general term:$T_{6+1} = ^9C_6 \left( \frac{3}{2} \right)^{9-6} \left( -\frac{1}{3} \right)^6$$T_7 = ^9C_3 \left( \frac{3}{2} \right)^3 \left( \frac{1}{3^6} \right)$$T_7 = ^9C_3 \cdot \frac{3^3}{2^3} \cdot \frac{1}{3^6} = ^9C_3 \cdot \frac{1}{2^3 \cdot 3^3} = ^9C_3 \cdot \frac{1}{6^3}$

In the expansion of ${\left( \frac{3x^2}{2} - \frac{1}{3x} \right)^9}$,the term independent of $x$ is

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