The expansion of (1+x+x^2)^{-3/2} in powers o | vedclass

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(C) The binomial expansion $(1+u)^n$ for $n 
otin N$ is valid if $|u| < 1$. Here,$u = x+x^2$. So,the expansion of $(1+x+x^2)^{-3/2}$ is valid if $|x^

Vedclass · Accepted Answer

$\left|x+\frac{1}{2}\right| < \frac{\sqrt{5}}{2}$

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(C) The binomial expansion $(1+u)^n$ for $n otin N$ is valid if $|u| Here,$u = x+x^2$. So,the expansion of $(1+x+x^2)^{-3/2}$ is valid if $|x^2+x| This implies $-1 Solving $x^2+x $x^2+x-1 The roots of $x^2+x-1 = 0$ are $x = \frac{-1 \pm \sqrt{1-4(1)(-1)}}{2} = \frac{-1 \pm \sqrt{5}}{2}$. Thus,$x^2+x-1 Also,$x^2+x > -1$ is always true since $x^2+x+1 = (x+\frac{1}{2})^2 + \frac{3}{4} > 0$. Combining these,we require $|x^2+x| < 1$,which is equivalent to $\left|x+\frac{1}{2} ight|^2 < \frac{5}{4}$,or $\left|x+\frac{1}{2} ight| < \frac{\sqrt{5}}{2}$.

The expansion of $(1+x+x^2)^{-3/2}$ in powers of $x$ is valid if

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