The coefficient of $\frac{1}{x}$ in the expansion of $(1 + x)^n (1 + \frac{1}{x})^n$ is

  • A
    $\frac{n!}{(n - 1)!(n + 1)!}$
  • B
    $\frac{(2n)!}{(n - 1)!(n + 1)!}$
  • C
    $\frac{n!}{(n - 1)!(n + 1)!}$
  • D
    None of these

Explore More

Similar Questions

If $\lim _{x \rightarrow 1} \frac{\sqrt[4]{x^{-3}}+a \sqrt[4]{x^5}}{x-1}=-2$,then the coefficient of $x$ in the expansion of $(\sqrt[4]{x^{-3}}+a \sqrt[4]{x^5})^4$ is

If the $k^{\text{th}}$ term in the expansion of $\left(\frac{3}{2} x^2 - \frac{1}{3x}\right)^6$ is independent of $x$,then the numerically greatest term in the expansion of $\left(\frac{3}{2} x^2 - \frac{1}{3x}\right)^k$ when $x = \frac{2}{3}$ is:

If $n$ is a positive integer and $f(n)$ is the coefficient of $x^n$ in the expansion of $(1+x)(1-x)^n$,then $f(2023) = $

If the middle term in the expansion of $(1+x)^{2n}$ is the greatest term,then $x$ lies in the interval

The coefficient of $x^{-5}$ in the binomial expansion of ${\left( {\frac{{x + 1}}{{{x^{2/3}} - {x^{1/3}} + 1}} - \frac{{x - 1}}{{x - {x^{1/2}}}}} \right)^{10}}$ where $x \ne 0, 1$,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