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# In the expansion of ${\left[ {\dfrac{{(1 + x)}}{{(1 - x)}}} \right]^2},$ the coefficient of ${x^n}$ will be:a. $4n$b. $4n - 3$c. $4n + 1$d. None of these

Last updated date: 16th Jul 2024
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Hint:First use the distributive property of power to open the square brackets and then use law of indices for negative powers then expand the both factors by binomial expansion and then multiply factors which will make exponent equals “n” and then add all of them to get the coefficient.

Complete step by step solution:
To find the coefficient of ${x^n}$ in the expansion of ${\left[ {\dfrac{{(1 + x)}}{{(1 - x)}}} \right]^2},$ we will first expand both the factors and then multiply them as follows
$= {\left[ {\dfrac{{(1 + x)}}{{(1 - x)}}} \right]^2} \\ = \dfrac{{{{(1 + x)}^2}}}{{{{(1 - x)}^2}}} \\$
We can write it as
$= {(1 + x)^2}{(1 - x)^{ - 2}}$
Now to expand the first factor we know the formula ${(a + b)^2} = {a^2} + 2ab + {b^2}$ and for second factor we have to use binomial theorem for negative powers which is given as
${(1 + x)^{ - n}} = 1 - nx + \dfrac{1}{2}n(n + 1){x^2} - \dfrac{1}{6}n(n + 1)(n + 2){x^3} + ....\;\;\;\;{\text{where}}\;\left| x \right| < 1$
Using both of these expansions to expand the above expression, we will get
$= \left( {1 + 2x + {x^2}} \right)\left[ {1 + 2x + 3{x^2} + .... + (n - 1){x^{n - 2}} + n{x^{n - 1}} + (n + 1){x^n} + ...} \right]$
Now if we see the multiplicands carefully, then we will get that there will be only three terms come from the multiplication that have coefficient ${x^n}$, the terms are as follows:
i. When $1$ present in the first multiplicand is multiplied with $(n + 1){x^n}$ in the second multiplicand.
ii. When $2x$ present in first multiplicand is multiplied with $n{x^{n - 1}}$
iii. And when ${x^2}$ present in first multiplicand is multiplied with $(n - 1){x^{n - 2}}$
So we will get the following terms after their multiplication,
$(n + 1){x^n},\;2n{x^n}\;and\;(n - 1){x^n}$ respectively
Therefore required coefficient of ${x^n}$ will be $n + 1 + 2n + n - 1 = 4n$

Note: Strictly note that binomial expansion for negative exponent formula is only applicable when the value of modulus of the variable is less than the constant present in it.
At last we have added all the coefficients of ${x^n}$ because terms having the same variable and exponent should be added to simplify the result.