The binomial expansion (7+3x)^{-2/5} is valid | vedclass

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Question

(B) Given the expression $(7+3x)^{-2/5} = 7^{-2/5}(1+\frac{3}{7}x)^{-2/5}$.Using the binomial expansion formula $(1+z)^n = 1 + nz + \frac{n(n-1)}{2!}z

Vedclass · Accepted Answer

$\frac{-108}{125}$

Vedclass · Answer

(B) Given the expression $(7+3x)^{-2/5} = 7^{-2/5}(1+\frac{3}{7}x)^{-2/5}$.Using the binomial expansion formula $(1+z)^n = 1 + nz + \frac{n(n-1)}{2!}z^2 + \frac{n(n-1)(n-2)}{3!}z^3 + \dots$,where $n = -2/5$ and $z = \frac{3x}{7}$.The $4^{th}$ term is given by $\frac{n(n-1)(n-2)}{3!} z^3$.Substituting the values:$T_4 = 7^{-2/5} 	imes \frac{(-2/5)(-2/5-1)(-2/5-2)}{3!} 	imes (\frac{3x}{7})^3$$T_4 = 7^{-2/5} 	imes \frac{(-2/5)(-7/5)(-12/5)}{6} 	imes \frac{27x^3}{343}$$T_4 = 7^{-2/5} 	imes \frac{-168/125}{6} 	imes \frac{27x^3}{7^3}$$T_4 = 7^{-2/5} 	imes (-\frac{28}{125}) 	imes \frac{27x^3}{7^3} = 7^{-2/5} 	imes (-\frac{756}{125 	imes 7^3}) x^3$$T_4 = 7^{-2/5} 	imes (-\frac{108}{125 	imes 7^2}) x^3 = -\frac{108}{125} 	imes 7^{-2/5-2} x^3 = -\frac{108}{125} 	imes 7^{-12/5} x^3$.Since $T_4 = kx^3$,we have $k = -\frac{108}{125} 	imes 7^{-12/5}$.Therefore,$7^{12/5}k = -\frac{108}{125}$.

The binomial expansion $(7+3x)^{-2/5}$ is valid for all $x$ in the interval $\left(\frac{-7}{3}, \frac{7}{3}\right)$. If the $4^{th}$ term of its expansion is $kx^3$,then the value of $(7^{12/5}k)$ is:

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