Q74 / 180 · COMEDK 2018 · Mathematics

If $x$ occurs is the expansion of $\left(x+\frac{1}{x}\right)^{n}$, then its coefficient is

  • A
    $\frac{n !}{(r !)^{1}}$
  • B
    $\frac{n !}{(r+1) !(r-1) !}$
  • Correct answer
    C
    $\frac{n !}{\left(\frac{n+r}{2}\right) !\left(\frac{n-r}{2}\right) !}$
  • D
    $\frac{n !}{\left[\left(\frac{n}{2}\right) !\right]^{2}}$

Solution

Let $x^{r}$ occurs in the expansion of $\left(x+\frac{1}{x}\right)^{n}$ $T_{r+1}={ }^{n} C_{p} x^{n-p}\left(\frac{1}{x}\right)^{p}$ ${ }^{n} C_{p} x^{n-p} x^{-p}={ }^{n} C_{p} x^{n-2 p}$ let $\quad n-2 p=r$ $\Rightarrow \quad n-r=2 p \Rightarrow p=\frac{n-r}{2}$ So, coefficient of $x^{r}={ }^{n} C_{p}=\frac{n !}{p !(n-p) !}$ $=\frac{n !}{\left(\frac{n-r}{2}\right) !\left(n-\frac{n-r}{2}\right) !}$ $=\frac{n !}{\left(\frac{n-r}{2}\right) !\left(\frac{n+r}{2}\right) !}$

Hence, the correct answer is option: 'C'.

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