If -frac{2}{3}

Question

(B) The expression is $(2-3x)^{-1/3} = 2^{-1/3} (1 - \frac{3x}{2})^{-1/3}$.Using the binomial expansion $(1-z)^{-n} = 1 + nz + \frac{n(n+1)}{2!}z^2 +

Vedclass · Accepted Answer

$\frac{35}{768(\sqrt[3]{2})}$

Vedclass · Answer

(B) The expression is $(2-3x)^{-1/3} = 2^{-1/3} (1 - \frac{3x}{2})^{-1/3}$.Using the binomial expansion $(1-z)^{-n} = 1 + nz + \frac{n(n+1)}{2!}z^2 + \frac{n(n+1)(n+2)}{3!}z^3 + \frac{n(n+1)(n+2)(n+3)}{4!}z^4 + \dots$,where $n = 1/3$ and $z = \frac{3x}{2}$.The $5^{	ext{th}}$ term is $T_5 = 2^{-1/3} 	imes \frac{n(n+1)(n+2)(n+3)}{4!} z^4$.Substituting $n = 1/3$:$T_5 = 2^{-1/3} 	imes \frac{(1/3)(4/3)(7/3)(10/3)}{24} 	imes (\frac{3x}{2})^4$.$T_5 = 2^{-1/3} 	imes \frac{280/81}{24} 	imes \frac{81x^4}{16} = 2^{-1/3} 	imes \frac{280}{24 	imes 16} x^4 = 2^{-1/3} 	imes \frac{35}{48} x^4$.Given $x = 1/2$,$x^4 = 1/16$.$T_5 = 2^{-1/3} 	imes \frac{35}{48} 	imes \frac{1}{16} = \frac{35}{768 	imes 2^{1/3}} = \frac{35}{768(\sqrt[3]{2})}$.

If $-\frac{2}{3} < x < \frac{2}{3}$,then the value of the $5^{\text{th}}$ term in the expansion of $\frac{1}{\sqrt[3]{2-3x}}$ when $x=\frac{1}{2}$ is

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