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

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(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

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

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(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} = $

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