If |x| 1,then (1 + x)^{-2} = | vedclass

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

Question

(D) Given that $|x| > 1$. Since $|x| > 1$,we have $|\frac{1}{x}| < 1$. We can rewrite the expression as: $(1 + x)^{-2} = [x(1 + \frac{1}{x})]^{-2} = x

Vedclass · Accepted Answer

$\frac{1}{x^2} - \frac{2}{x^3} + \frac{3}{x^4} - \dots$

Vedclass · Answer

(D) Given that $|x| > 1$. Since $|x| > 1$,we have $|\frac{1}{x}| We can rewrite the expression as: $(1 + x)^{-2} = [x(1 + \frac{1}{x})]^{-2} = x^{-2}(1 + \frac{1}{x})^{-2}$. Using the binomial expansion $(1 + z)^{-n} = 1 - nz + \frac{n(n+1)}{2!}z^2 - \dots$ for $|z| $x^{-2}(1 + \frac{1}{x})^{-2} = \frac{1}{x^2} [1 - 2(\frac{1}{x}) + \frac{2(3)}{2!}(\frac{1}{x})^2 - \dots]$ $= \frac{1}{x^2} [1 - \frac{2}{x} + \frac{3}{x^2} - \dots]$ $= \frac{1}{x^2} - \frac{2}{x^3} + \frac{3}{x^4} - \dots$

If $|x| > 1$,then $(1 + x)^{-2} = $

Similar Questions

Assuming $x$ to be so small that $x^2$ and higher powers of $x$ can be neglected,the coefficient of $x$ in $\frac{(1-x)^{1/3}+(1-5x)^2}{(16-x)^{1/4}}$ is equal to

$1 - \frac{3}{16} + \frac{1 \cdot 4}{1 \cdot 2} \left(\frac{3}{16}\right)^2 - \frac{1 \cdot 4 \cdot 7}{1 \cdot 2 \cdot 3} \left(\frac{3}{16}\right)^3 + \ldots$

If $|x| < \frac{1}{2}$,then the coefficient of $x^r$ in the expansion of $\frac{1+2x}{(1-2x)^2}$ is

If $n$ is a positive integer and the coefficient of $x^{10}$ in the expansion of $(1+x)^{15}$ is equal to the coefficient of $x^5$ in the expansion of $(1-x)^{-n}$,then $n=$

When $|x|>3$,the coefficient of $\frac{1}{x^n}$ in the expansion of $x^{3/2}(3+x)^{1/2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module