Question

What is the coefficient of \[\dfrac{1}{x}\] in the expansion \[{(1 + x)^n}{\left( {1 + \dfrac{1}{x}} \right)^n}\].
(a). \[\dfrac{{n!}}{{(n - 1)!(n + 1)!}}\]
(b). \[\dfrac{{2n!}}{{(n - 1)!(n + 1)!}}\]
(c). \[\dfrac{{2n!}}{{(2n - 1)!(2n + 1)!}}\]
(d). None of these

Answer 1

Hint: Simplify the product \[{(1 + x)^n}{\left( {1 + \dfrac{1}{x}} \right)^n}\] and obtain an expression of a single exponent of a sum term. Then, use the nth term of the binomial expansion formula to find the coefficient of \[\dfrac{1}{x}\].

Complete step-by-step answer:
We need to find the coefficient of \[\dfrac{1}{x}\] in the expansion \[{(1 + x)^n}{\left( {1 + \dfrac{1}{x}} \right)^n}\]. It is not possible to multiply all terms to find the value of the coefficient.
Let us simplify the expansion \[{(1 + x)^n}{\left( {1 + \dfrac{1}{x}} \right)^n}\] to contain only one binomial term.
We can write \[1 + \dfrac{1}{x}\] as \[\dfrac{{1 + x}}{x}\], then, we have the following:
\[{(1 + x)^n}{\left( {1 + \dfrac{1}{x}} \right)^n} = {(1 + x)^n}{\left( {\dfrac{{1 + x}}{x}} \right)^n}\]
Taking \[{x^n}\] in the denominator common outside, we have multiplication of \[{(1 + x)^n}\] with itself, which is \[{(1 + x)^{2n}}\].
\[{(1 + x)^n}{\left( {1 + \dfrac{1}{x}} \right)^n} = \dfrac{1}{{{x^n}}}{(1 + x)^n}{(1 + x)^n}\]
\[{(1 + x)^n}{\left( {1 + \dfrac{1}{x}} \right)^n} = \dfrac{1}{{{x^n}}}{(1 + x)^{2n}}\]
Hence, we now have a single binomial term.
We can easily find the coefficient of \[\dfrac{1}{x}\] from the expansion \[\dfrac{1}{{{x^n}}}{(1 + x)^{2n}}\].
It is equal to the coefficient of \[{x^{n - 1}}\] in the expansion \[{(1 + x)^{2n}}\].
We know that the coefficient of \[{x^r}\] of a binomial expansion \[{(1 + x)^{2n}}\] is the (r + 1)th term and is given as follows:
\[{T_{r + 1}} = {}^{2n}{C_r}\]
We have to find the coefficient of the \[{x^{n - 1}}\], which is the nth term in the expansion \[{(1 + x)^{2n}}\] is given as follows:
\[{T_n} = {}^{2n}{C_{n - 1}}............(1)\]
  We know the formula for \[{}^n{C_r}\] is given as follows:
\[{}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}.............(2)\]
Using formula (2) in equation (1), we have:
\[{}^{2n}{C_{n - 1}} = \dfrac{{2n!}}{{(n - 1)!(2n - (n - 1))!}}\]
Simplifying, we get:
\[{}^{2n}{C_{n - 1}} = \dfrac{{2n!}}{{(n - 1)!(n + 1)!}}\]
Hence, the correct answer is option (b).

Note: You can also expand the terms \[{(1 + x)^n}\] and \[{\left( {1 + \dfrac{1}{x}} \right)^n}\]. Then, find the sum of the terms of the coefficients of \[\dfrac{1}{x}\] and simplify them using formulas for combination.