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Question: What is the coefficient of \[\dfrac{1}{x}\] in the expansion \[{(1 + x)^n}{\left( {1 + \dfrac{1}{x}}...

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

Explanation

Solution

Hint: Simplify the product (1+x)n(1+1x)n{(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 1x\dfrac{1}{x}.

Complete step-by-step answer:
We need to find the coefficient of 1x\dfrac{1}{x} in the expansion (1+x)n(1+1x)n{(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(1+1x)n{(1 + x)^n}{\left( {1 + \dfrac{1}{x}} \right)^n} to contain only one binomial term.
We can write 1+1x1 + \dfrac{1}{x} as 1+xx\dfrac{{1 + x}}{x}, then, we have the following:
(1+x)n(1+1x)n=(1+x)n(1+xx)n{(1 + x)^n}{\left( {1 + \dfrac{1}{x}} \right)^n} = {(1 + x)^n}{\left( {\dfrac{{1 + x}}{x}} \right)^n}
Taking xn{x^n} in the denominator common outside, we have multiplication of (1+x)n{(1 + x)^n} with itself, which is (1+x)2n{(1 + x)^{2n}}.
(1+x)n(1+1x)n=1xn(1+x)n(1+x)n{(1 + x)^n}{\left( {1 + \dfrac{1}{x}} \right)^n} = \dfrac{1}{{{x^n}}}{(1 + x)^n}{(1 + x)^n}
(1+x)n(1+1x)n=1xn(1+x)2n{(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 1x\dfrac{1}{x} from the expansion 1xn(1+x)2n\dfrac{1}{{{x^n}}}{(1 + x)^{2n}}.
It is equal to the coefficient of xn−1{x^{n - 1}} in the expansion (1+x)2n{(1 + x)^{2n}}.
We know that the coefficient of xr{x^r} of a binomial expansion (1+x)2n{(1 + x)^{2n}} is the (r + 1)th term and is given as follows:
Tr+1=2nCr{T_{r + 1}} = {}^{2n}{C_r}
We have to find the coefficient of the xn−1{x^{n - 1}}, which is the nth term in the expansion (1+x)2n{(1 + x)^{2n}} is given as follows:
Tn=2nCn−1............(1){T_n} = {}^{2n}{C_{n - 1}}............(1)
We know the formula for nCr{}^n{C_r} is given as follows:
nCr=n!r!(n−r)!.............(2){}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}.............(2)
Using formula (2) in equation (1), we have:
2nCn−1=2n!(n−1)!(2n−(n−1))!{}^{2n}{C_{n - 1}} = \dfrac{{2n!}}{{(n - 1)!(2n - (n - 1))!}}
Simplifying, we get:
2nCn−1=2n!(n−1)!(n+1)!{}^{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{(1 + x)^n} and (1+1x)n{\left( {1 + \dfrac{1}{x}} \right)^n}. Then, find the sum of the terms of the coefficients of 1x\dfrac{1}{x} and simplify them using formulas for combination.