The correct matching of List-$I$ from List-$II$ is:
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 $(iv)$ $\frac{x}{x-1}$
  $(v)$ $\frac{x^4}{(x^2+1)^2}$
  $(vi)$ $\frac{x^4}{(x^2-1)^2}$

  • A
    $(A)-(i), (B)-(iii), (C)-(iv), (D)-(v)$
  • B
    $(A)-(ii), (B)-(iii), (C)-(iv), (D)-(v)$
  • C
    $(A)-(iii), (B)-(ii), (C)-(iv), (D)-(v)$
  • D
    $(A)-(ii), (B)-(iii), (C)-(i), (D)-(v)$

Explore More

Similar Questions

$1+\frac{4}{15}+\frac{4 \times 10}{15 \times 30}+\frac{4 \times 10 \times 16}{15 \times 30 \times 45}+\ldots \quad \infty=$

$1 - \frac{1}{8} + \frac{1}{8} \cdot \frac{3}{16} - \frac{1 \cdot 3 \cdot 5}{8 \cdot 16 \cdot 24} + \dots =$

Difficult
View Solution

$\frac{1}{8} - \frac{7}{8 \times 12} + \frac{7 \times 10}{8 \times 12 \times 16} - \ldots =$

$1 + \frac{1}{4} + \frac{1 \times 3}{4 \times 8} + \frac{1 \times 3 \times 5}{4 \times 8 \times 12} + \dots = $

Difficult
View Solution

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

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo