Question

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

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

Given expansion is ${(1+x)}^n{(1-\frac{1}{x})}^n={(x-\frac{1}{x})}^n$,

The general term in the expansion is ${(-1)}^r\left(\begin{array}{c}n\\ r\end{array}\right)x^{n-r}{\left(\frac{1}{x}\right)}^r=\left(\begin{array}{c}n\\ r\end{array}\right)x^{n-2r}$

So to find the term independent of $x$

$⟹r=\frac{n}{2}$,

When $n$ is odd the combination doesnt exist so the term independent of $x$ is $0$,

When $n$ is even the term independent of $x$ is ${(-1)}^r\left(\begin{array}{c}n\\ r\end{array}\right)$