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Question: The coefficient of \({x^n}\) in expansion of\((1 + x){(1 - x)^n}\) is A.\(n - 1\) B.\({( - 1)^{n...

The coefficient of xn{x^n} in expansion of(1+x)(1−x)n(1 + x){(1 - x)^n} is
A.n−1n - 1
B.(−1)n−1n{( - 1)^{n - 1}}n
C.(−1)n−1(n−1)2{( - 1)^{n - 1}}{(n - 1)^2}
D.(−1)n(1−n){( - 1)^n}(1 - n)

Explanation

Solution

Hint : Here before solving this question we need to know the following formula to solve question: -
Binomial theorem
(a+b)n=nC0anb0+nC1an−1b1+nC2an−2b2+......+nCnbn{(a + b)^n} = {}^n{C_0}{a^n}{b^0} + {}^n{C_1}{a^{n - 1}}{b^1} + {}^n{C_2}{a^{n - 2}}{b^2} + ...... + {}^n{C_n}{b^n}
Since we know that combination can be written as,
nCr=n!r!(n−r)!{}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}

Complete step-by-step answer :
Now, the given equation is,
(1+x)(1−x)n(1 + x){(1 - x)^n}
Using binomial theorem on (1−x)n{(1 - x)^n} we get

(1−x)n=nC0(1)n+nC1(1)n−1(−x)+nC2(1)n−2(−x)2+nC3(1)n−3(−x)3+  ...... +nCn−1(1)(−x)n−1+nCn(−x)n (1−x)n=nC0+nC1(−x)+nC2(−x)2+nC3(−x)3+  ...... +nCn−1(−x)n−1+nCn(−x)n  {(1 - x)^n} = {}^n{C_0}{(1)^n} + {}^n{C_1}{(1)^{n - 1}}( - x) + {}^n{C_2}{(1)^{n - 2}}{( - x)^2} + {}^n{C_3}{(1)^{n - 3}}{( - x)^3} + \,\,......\, + {}^n{C_{n - 1}}(1){( - x)^{n - 1}} + {}^n{C_n}{( - x)^n} \\\ {(1 - x)^n} = {}^n{C_0} + {}^n{C_1}( - x) + {}^n{C_2}{( - x)^2} + {}^n{C_3}{( - x)^3} + \,\,......\, + {}^n{C_{n - 1}}{( - x)^{n - 1}} + {}^n{C_n}{( - x)^n} \\\

Writing combination formula

(1−x)n=n!0!(n−0)!+n!1!(n−1)!(−x)+n!2!(n−2)!(x2)+n!3!(n−3)!(−x3)+ ......+n!(n−1)!1!(−x)n−1 \+n!(n−n)!0!(−x)n  {(1 - x)^n} = \dfrac{{n!}}{{0!\left( {n - 0} \right)!}} + \dfrac{{n!}}{{1!\left( {n - 1} \right)!}}( - x) + \dfrac{{n!}}{{2!\left( {n - 2} \right)!}}({x^2}) + \dfrac{{n!}}{{3!\left( {n - 3} \right)!}}( - {x^3}) + \,...... + \dfrac{{n!}}{{(n - 1)!1!}}{( - x)^{n - 1}} \\\ \+ \dfrac{{n!}}{{\left( {n - n} \right)!0!}}{( - x)^n} \\\

Further simplifying
(1−x)n=1−nx+n(n−1)2x2−n(n−1)(n−2)6x3+ ....... +n(−x)n−1+(−x)n{(1 - x)^n} = 1 - nx + \dfrac{{n(n - 1)}}{2}{x^2} - \dfrac{{n(n - 1)(n - 2)}}{6}{x^3} + \,.......\, + n{( - x)^{n - 1}} + {( - x)^n}
Further given equation can be simplified as,

(1+x)(1−x)n=(1+x)(1−nx+n(n−1)2x2−n(n−1)(n−2)6x3+ ....... +n(−x)n−1+(−x)n)(1 + x){(1 - x)^n} = (1 + x)(1 - nx + \dfrac{{n(n - 1)}}{2}{x^2} - \dfrac{{n(n - 1)(n - 2)}}{6}{x^3} + \,.......\, + n{( - x)^{n - 1}} + {( - x)^n})
See, now we consider only that terms that contain xn{x^n} and that is
1×(−x)n+x×n(−x)n−1 (−x)n(1−n) (−1)n(1−n)xn  1 \times {( - x)^n} + x \times n{( - x)^{n - 1}} \\\ {( - x)^n}(1 - n) \\\ {( - 1)^n}(1 - n){x^n} \\\
Thus, we found that the coefficient is (−1)n(1−n){( - 1)^n}(1 - n)
So, the correct answer is “Option D”.

Note : In question of binomial expansion there are two major sources of concern.
When there is a negative operator the odd exponents will have negative coefficient. While substituting the formula we must not get confused with the permutation formula as there is very little difference between the two.