If |x|

Question

(A) The binomial expansion for a negative index is given by $(1 - y)^{-n} = 1 + ny + \frac{n(n + 1)}{2!}y^2 + \dots \infty$.Comparing this with the gi

Vedclass · Accepted Answer

$\left( \frac{1 + x}{1 - x} \right)^n$

Vedclass · Answer

(A) The binomial expansion for a negative index is given by $(1 - y)^{-n} = 1 + ny + \frac{n(n + 1)}{2!}y^2 + \dots \infty$.Comparing this with the given series $1 + n\left( \frac{2x}{1 + x} \right) + \frac{n(n + 1)}{2!}\left( \frac{2x}{1 + x} \right)^2 + \dots \infty$,we identify $y = \frac{2x}{1 + x}$.Thus,the sum is $\left( 1 - \frac{2x}{1 + x} \right)^{-n}$.Simplifying the expression inside the bracket: $1 - \frac{2x}{1 + x} = \frac{1 + x - 2x}{1 + x} = \frac{1 - x}{1 + x}$.Therefore,the sum is $\left( \frac{1 - x}{1 + x} \right)^{-n} = \left( \frac{1 + x}{1 - x} \right)^n$.

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	$(iv)$ $\frac{x}{x-1}$
	$(v)$ $\frac{x^4}{(x^2+1)^2}$
	$(vi)$ $\frac{x^4}{(x^2-1)^2}$

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

Similar Questions

If $|x| < 1$,then the number of terms in the expansion of $[\frac{1}{2}(1 \cdot 2 + 2 \cdot 3 x + 3 \cdot 4 x^2 + . . . . . . \infty)]^{-25}$ is

$\frac{1}{4}-\frac{5}{4 \cdot 8}+\frac{5 \cdot 7}{4 \cdot 8 \cdot 12}-\ldots=$

In the expansion of $\frac{2x+1}{(1+x)(1-2x)}$,the sum of the coefficients of the first $5$ odd powers of $x$ is

The expansion of $\frac{1}{\sqrt{4 - 3x}}$ using the binomial theorem is valid if:

Vedclass Test Series

Exam Paper Generator

Online Exam Module