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

Similar Questions

The formula $(a + b)^m = a^m + ma^{m-1}b + \frac{m(m - 1)}{1 \cdot 2}a^{m - 2}b^2 + \dots$ holds when

$\frac{1}{4}-\frac{5}{4 \cdot 8}+\frac{5 \cdot 7}{4 \cdot 8 \cdot 12}-\ldots=$

If $x = \frac{2 \cdot 5}{(2!) 3} + \frac{2 \cdot 5 \cdot 7}{(3!) 3^2} + \frac{2 \cdot 5 \cdot 7 \cdot 9}{(4!) 3^3} + \dots$,then $x^2 + 8x + 8 = $

If $y = \frac{3}{4} + \frac{3 \cdot 5}{4 \cdot 8} + \frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12} + \dots \infty$,then

If $x = \frac{1}{5} + \frac{1 \times 3}{5 \times 10} + \frac{1 \times 3 \times 5}{5 \times 10 \times 15} + \ldots$,then $3x^2 + 6x =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module