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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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:

Similar Questions

$\frac{1}{8} - \frac{7}{8 \times 12} + \frac{7 \times 10}{8 \times 12 \times 16} - \ldots =$

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

If $x$ is small,so that $x^2$ and higher powers can be neglected,then the approximate value for $\frac{(1-2 x)^{-1}(1-3 x)^{-2}}{(1-4 x)^{-3}}$ is

If $|x| < 1$,then the value of $1 + n\left( \frac{2x}{1 + x} \right) + \frac{n(n + 1)}{2!}\left( \frac{2x}{1 + x} \right)^2 + \dots \infty$ will be

The coefficient of $x^3$ in the expansion of $(1-2x)^{1/2}(1+3x)^{-1/3}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module