Question

The term independent of $x$ in the expansion of $(1+x{)}^n.{(1-\frac{1}{x})}^n$ if $(n\in N)$, is

• A. $0$, if $n$ is odd
• B. $(-1{)}^{\frac{n-2}{2}}{.}^n{C}_{\frac{n-1}{2}},$ if $n$ is odd
• C. $(-1{)}^{\frac{n}{2}}{.}^n{C}_{\frac{n}{2}},$ if $n$ is even
• D. ${}^n{C}_{\frac{n}{2}},$ if $n$ is even

Answer 1

Answer: (A), (C)

The correct options are
A $0$, if $n$ is odd
C $(-1{)}^{\frac{n}{2}}{.}^n{C}_{\frac{n}{2}},$ if $n$ is even

$(x+1{)}^n(1-\frac{1}{x}{)}^n$
$=(x+1{)}^n(\frac{x-1}{x}{)}^n$
$=(\frac{x^2-1}{x}{)}^n$
$=(x-\frac{1}{x}{)}^n$
$T_{r+1}=(-1{)}^r{}^n{C}_r{x}^{n-2r}$
If $n$ is even then $r=n/2$.
Hence coefficient of term independent of $x$ will be${}^n{C}_{\frac{n}{2}}$.
If $n$ is odd then there won't be any term independent of $x$.