The coefficient of x^6 in the binomial expans | vedclass

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(B) The general term $T_{r+1}$ in the expansion of ${\left( \frac{4x^2}{3} - \frac{3}{2x} \right)^9}$ is given by:$T_{r+1} = {^9C_r} {\left( \frac{4x^

Vedclass · Accepted Answer

$2688$

Vedclass · Answer

(B) The general term $T_{r+1}$ in the expansion of ${\left( \frac{4x^2}{3} - \frac{3}{2x} ight)^9}$ is given by:$T_{r+1} = {^9C_r} {\left( \frac{4x^2}{3} ight)^{9-r}} {\left( -\frac{3}{2x} ight)^r}$$T_{r+1} = {^9C_r} {\left( \frac{4}{3} ight)^{9-r}} {\left( -\frac{3}{2} ight)^r} {x^{2(9-r)}} {x^{-r}}$$T_{r+1} = {^9C_r} {\left( \frac{4}{3} ight)^{9-r}} {\left( -\frac{3}{2} ight)^r} {x^{18-3r}}$To find the coefficient of $x^6$,we set the exponent of $x$ equal to $6$:$18 - 3r = 6$$3r = 12 \Rightarrow r = 4$Now,substitute $r = 4$ into the expression for the coefficient:Coefficient $= {^9C_4} {\left( \frac{4}{3} ight)^{9-4}} {\left( -\frac{3}{2} ight)^4}$$= \frac{9 	imes 8 	imes 7 	imes 6}{4 	imes 3 	imes 2 	imes 1} 	imes {\left( \frac{4}{3} ight)^5} 	imes {\left( \frac{3}{2} ight)^4}$$= 126 	imes \frac{4^5}{3^5} 	imes \frac{3^4}{2^4}$$= 126 	imes \frac{(2^2)^5}{3^5} 	imes \frac{3^4}{2^4}$$= 126 	imes \frac{2^{10}}{2^4} 	imes \frac{3^4}{3^5}$$= 126 	imes 2^6 	imes \frac{1}{3}$$= 42 	imes 64 = 2688$

The coefficient of $x^6$ in the binomial expansion of ${\left( \frac{4x^2}{3} - \frac{3}{2x} \right)^9}$ is

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