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

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

Question

(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

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

Vedclass · Answer

(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 $\left|x+\frac{1}{2} ight| < \frac{\sqrt{5}}{2}$.

List-$I$	List-$II$
$(A)$ $(1-x)^{-n}$	$(i)$ $\frac{x}{x+1}$
$(B)$ $(1+x)^{-n}$	$(ii)$ $1-nx+\frac{n(n+1)}{2!}x^2-\dots$ if $\|x\| < 1$
$(C)$ If $x>1$,then $1+\frac{1}{x}+\frac{1}{x^2}+\dots$ is	$(iii)$ $1+nx+\frac{n(n+1)}{2!}x^2+\dots$ if $\|x\| < 1$
$(D)$ If $\|x\|>1$,then $1-\frac{2}{x^2}+\frac{3}{x^4}-\frac{4}{x^6}+\dots$ is	$(iv)$ $\frac{x}{x-1}$
	$(v)$ $\frac{x^4}{(x^2+1)^2}$
	$(vi)$ $\frac{x^4}{(x^2-1)^2}$

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

Similar Questions

The sum of the series $1+\frac{2}{3}\left(\frac{1}{8}\right)+\frac{2 \times 5}{3 \times 6}\left(\frac{1}{8}\right)^2+\frac{2 \times 5 \times 8}{3 \times 6 \times 9}\left(\frac{1}{8}\right)^3+\ldots$ is

$\frac{1}{4}-\frac{5}{4 \cdot 8}+\frac{5 \cdot 9}{4 \cdot 8 \cdot 12}-\ldots=$

If $(a+bx)^{-3} = \frac{1}{27} + \frac{1}{3}x + \dots$,then the ordered pair $(a, b)$ is equal to

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module