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Question: The coefficient of \(\dfrac{1}{x}\) in the expansion of \({(1 + x)^n}{\left( {1 + \dfrac{1}{x}} \rig...

The coefficient of 1x\dfrac{1}{x} in the expansion of (1+x)n(1+1x)n{(1 + x)^n}{\left( {1 + \dfrac{1}{x}} \right)^n} is
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

According to given in the question we have to find the coefficient of 1x\dfrac{1}{x} in the expansion of (1+x)n(1+1x)n{(1 + x)^n}{\left( {1 + \dfrac{1}{x}} \right)^n}. So, first of all we have to simplify the expression given in the question so, we have to determine the L.C.M and then we have to multiply all the terms of the expression to simplify the expression.
Now, to solve the expansion we have to use the formula as mentioned below:

Formula used: ⇒Tr+1=cr2nxr−n.....................(A) \Rightarrow {T_{r + 1}} = c_r^{2n}{x^{r - n}}.....................(A)
Now, we have to determine this for 1x\dfrac{1}{x} with the help of the formula above.
Hence, after determining r we have to substitute and the values in the formula and then we have to determine the coefficient of 1x\dfrac{1}{x} or x−1{x^{ - 1}}

Complete step-by-step solution:
Step 1: First of all we have to simplify the given algebraic expression by determining the L.C.M of the terms and after that we have to multiply the terms with each other as mentioned in the solution hint. Hence,

=(1+x)n(1+x)nxn =(1+x)2nxn...............(1) = {(1 + x)^n}\dfrac{{{{(1 + x)}^n}}}{{{x^n}}} \\\ = \dfrac{{{{(1 + x)}^{2n}}}}{{{x^n}}}...............(1)

Step 2: Now, we have to rearrange the terms of the algebraic expression (1) as obtained in the solution step 1.
=1xn[(1+x)n]= \dfrac{1}{{{x^n}}}\left[ {{{\left( {1 + x} \right)}^n}} \right]…………………..(2)
Step 3: Now, to solve the expression (2) as obtained in the solution step 2 we have to use the formula (A) as mentioned in the solution hint. Hence,
⇒cr2nxr−n=x−1\Rightarrow c_r^{2n}{x^{r - n}} = {x^{ - 1}}……………….(3)
Step 4: Now, to obtain the coefficient of 1x\dfrac{1}{x}or x−1{x^{ - 1}}we have to compare as in the expression (3) as obtained in the solution step 3.
⇒r−n=−1 ⇒r=n−1 \Rightarrow r - n = - 1 \\\ \Rightarrow r = n - 1
Step 5: Now, on substituting the value of r in the formula (A) to find the coefficient hence,
⇒cn−12n=(2n)!(n+1)!(n−1)!\Rightarrow c_{n - 1}^{2n} = \dfrac{{(2n)!}}{{(n + 1)!(n - 1)!}}
Final solution: Hence, with the help of the formula (A) we have obtained the coefficient of 1x\dfrac{1}{x}in the expansion of (1+x)n(1+1x)n{(1 + x)^n}{\left( {1 + \dfrac{1}{x}} \right)^n} is (2n)!(n+1)!(n−1)!\dfrac{{(2n)!}}{{(n + 1)!(n - 1)!}}.

Therefore option (B) is correct.

Note: Coefficient is a multiplicative factor in some polynomial, any expression, or any series; it is usually a number, but may be any expression.
Some specific coefficients that occur frequently in mathematics have dedicated names. As the binomial in the expand form of (x+y)n{(x + y)^n}