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

The coefficient of xn{x^n} in expansion of (1+x)(1−x)n(1 + x){(1 - x)^n} is
(A) (n-1)
(B)(−1)n(1−n){( - 1)^n}(1 - n)
(C) (−1)n−1(n−1)2{( - 1)^{n - 1}}{(n - 1)^2}
(D)(−1)n−1n{( - 1)^{n - 1}}n
If the coefficient of rth,(r+1)thand(r+2)th{r^{th}},{(r + 1)^{th}}and{(r + 2)^{th}} terms in the binomial expansion (1+y)m{(1 + y)^m} are in A.P, then, m and r satisfy the equation-
(A)m2−m(4r−1)+4r2+2=0{m^2} - m(4r - 1) + 4{r^2} + 2 = 0
(B) m2−m(4r+1)+4r2−2=0{m^2} - m(4r + 1) + 4{r^2} - 2 = 0
(C) m2−m(4r+1)+4r2+2=0{m^2} - m(4r + 1) + 4{r^2} + 2 = 0
(D) m2−m(4r−1)+4r2−2=0{m^2} - m(4r - 1) + 4{r^2} - 2 = 0

Explanation

Solution

Here we are going to find the coefficient of xn{x^n} using the expansion of the given binomial equation.
Initially we will expand the binomial expansion and multiply it with (1+x) and then find the final coefficient required.

Formula used: The Binomial Theorem states that, where n is a positive integer,
(x+y)n=∑k=0nnCkxn−kyk=xn+nC1xn−1y+nC2xn−2y2+......{(x + y)^n} = \sum\limits_{k = 0}^n {^n{C_k}{x^{n - k}}{y^k}} = {x^n}{ + ^n}{C_1}{x^{n - 1}}y{ + ^n}{C_2}{x^{n - 2}}{y^2} + ......

Complete step-by-step answer:
We have to find out the coefficient of xn{x^n} in expansion of(1+x)(1−x)n(1 + x){(1 - x)^n} .
From Binomial theorem we have,
(x+y)n=∑k=0nnCkxn−kyk=xn+nC1xn−1y+nC2xn−2y2+......{(x + y)^n} = \sum\limits_{k = 0}^n {^n{C_k}{x^{n - k}}{y^k}} = {x^n}{ + ^n}{C_1}{x^{n - 1}}y{ + ^n}{C_2}{x^{n - 2}}{y^2} + ......
Let us put x=1, y=-x in the above equation and apply binomial theorem to get the expansion of (1−x)n{(1 - x)^n}
(1−x)n=∑k=0nnCk(1)n−k(−x)k{(1 - x)^n} = \sum\limits_{k = 0}^n {^n{C_k}{{(1)}^{n - k}}{{( - x)}^k}}
And on expanding the summation in the above equation we get,(1−x)n=1+nC1(1)n−1(−x)+nC2(1)n−2(−x)2+......+nCn−2(1)n−(n−2)(−x)n−2+nCn−1(1)n−(n−1)(−x)n−1+(−x)n{(1 - x)^n} = 1{ + ^n}{C_1}{(1)^{n - 1}}( - x){ + ^n}{C_2}{(1)^{n - 2}}{( - x)^2} + ......{ + ^n}{C_{n - 2}}{(1)^{n - (n - 2)}}{( - x)^{n - 2}}{ + ^n}{C_{n - 1}}{(1)^{n - (n - 1)}}{( - x)^{n - 1}} + {( - x)^n}Thus we can see from expansion of (1+x)(1−x)n(1 + x){(1 - x)^n}
(1+x)(1−x)n=1.(1−x)n+x(1−x)n(1 + x){(1 - x)^n} = 1.{(1 - x)^n} + x{(1 - x)^n}
Let us use the binomial expansion so that we get,

(1+x)(1−x)n =1.1+nC1(1)n−1(−x)+nC2(1)n−2(−x)2+......+nCn−2(1)n−(n−2)1.(−x)n−2+nCn−1(1)n−(n−1)1.(−x)n−1+(−x)n \+x1+nC1(1)n−1(−x)+nC2(1)n−2(−x)2+......+nCn−2(1)n−(n−2)1.(−x)n−2+nCn−1(1)n−(n−1)1.(−x)n−1+(−x)n  (1 + x){(1 - x)^n} \\\ = 1.\\{ 1{ + ^n}{C_1}{(1)^{n - 1}}( - x){ + ^n}{C_2}{(1)^{n - 2}}{( - x)^2} + ......{ + ^n}{C_{n - 2}}{(1)^{n - (n - 2)}}1.{( - x)^{n - 2}}{ + ^n}{C_{n - 1}}{(1)^{n - (n - 1)}}1.{( - x)^{n - 1}} + {( - x)^n}\\} \\\ \+ x\\{ 1{ + ^n}{C_1}{(1)^{n - 1}}( - x){ + ^n}{C_2}{(1)^{n - 2}}{( - x)^2} + ......{ + ^n}{C_{n - 2}}{(1)^{n - (n - 2)}}1.{( - x)^{n - 2}}{ + ^n}{C_{n - 1}}{(1)^{n - (n - 1)}}1.{( - x)^{n - 1}} + {( - x)^n}\\} \\\

The coefficient of xn{x^n} in expansion of (1+x)(1−x)n(1 + x){(1 - x)^n}is
(−1)n+nCn−1(1)n−(n−1)1.(−1)n−1{( - 1)^n}{ + ^n}{C_{n - 1}}{(1)^{n - (n - 1)}}1.{( - 1)^{n - 1}}
Let us solve using the combination formula we get,
(−1)n+nCn−1(1)n−(n−1)1.(−1)n−1=(−1)n+n!(n−1)!(n−n+1)!1.(−1)n−1{( - 1)^n}{ + ^n}{C_{n - 1}}{(1)^{n - (n - 1)}}1.{( - 1)^{n - 1}} = {( - 1)^n} + \dfrac{{n!}}{{(n - 1)!(n - n + 1)!}}1.{( - 1)^{n - 1}}
On further simplifications using factorial we get
(−1)n+n(n−1)!(n−1)!.11.(−1)n−1=(−1)n−1(−1+n){( - 1)^n} + \dfrac{{n(n - 1)!}}{{(n - 1)!.1}}1.{( - 1)^{n - 1}} = {( - 1)^{n - 1}}( - 1 + n)
We further modify it to find the answer,
(−1)n+nCn−1(1)n−(n−1)1.(−1)n−1=(−1)n(1−n){( - 1)^n}{ + ^n}{C_{n - 1}}{(1)^{n - (n - 1)}}1.{( - 1)^{n - 1}} = {( - 1)^n}(1 - n)
Hence, (B) is the correct option.

We have to find out the coefficient of rth,(r+1)thand(r+2)th{r^{th}},{(r + 1)^{th}}and{(r + 2)^{th}} terms in the binomial expansion(1+y)m{(1 + y)^m}.
From Binomial theorem we have,
(x+y)n=∑k=0nnCkxn−kyk=xn+nC1xn−1y+nC2xn−2y2+......{(x + y)^n} = \sum\limits_{k = 0}^n {^n{C_k}{x^{n - k}}{y^k}} = {x^n}{ + ^n}{C_1}{x^{n - 1}}y{ + ^n}{C_2}{x^{n - 2}}{y^2} + ......
Let us put x=1, n=m and apply binomial theorem to get the expansion of (1+y)m{(1 + y)^m}

{(1 + y)^m} = \sum\limits_{k = 0}^m {^m{C_k}{1^{m - k}}{y^k}} \\\ = 1{ + ^m}{C_1}{1^{m - 1}}y{ + ^m}{C_2}{1^{m - 2}}{y^2} + {......^m}{C_{r - 1}}{1^{m - (r - 1)}}{y^{r - 1}}{ + ^m}{C_r}{1^{m - r}}{y^r}{ + ^m}{C_{r + 1}}{1^{m - (r + 1)}}{y^{r + 1}} + .... \\\ \end{gathered} $$ The coefficient of $${r^{th}},{(r + 1)^{th}}and{(r + 2)^{th}}$$ terms in the binomial expansion $${(1 + y)^m}$$are $$^m{C_{r - 1}}{,^m}{C_r}{,^m}{C_{r + 1}}$$ It is given that, $${r^{th}},{(r + 1)^{th}}and{(r + 2)^{th}}$$ terms in the binomial expansion $${(1 + y)^m}$$are in A.P So, we get, $$\dfrac{{^m{C_{r + 1}}{ + ^m}{C_{r - 1}}}}{2}{ = ^m}{C_r}$$ Rearrange the above equation we get, $$\dfrac{{^m{C_{r + 1}}{ + ^m}{C_{r - 1}}}}{{^m{C_r}}} = 2$$ Let us use the combination formula so that we get, $$\dfrac{{\dfrac{{m!}}{{(r + 1)!(m - r - 1)!}}}}{{\dfrac{{m!}}{{r!(m - r)!}}}} + \dfrac{{\dfrac{{m!}}{{(r - 1)!(m - r + 1)!}}}}{{\dfrac{{m!}}{{r!(m - r)!}}}} = 2$$ Further solving of the above equation will lead to$$\dfrac{{m!}}{{(r + 1)!(m - r - 1)!}} \times \dfrac{{r!(m - r)!}}{{m!}} + \dfrac{{m!}}{{(r - 1)!(m - r + 1)!}} \times \dfrac{{r!(m - r)!}}{{m!}} = 2$$ On solving the factorials we get, $$\dfrac{{m - r}}{{r + 1}} + \dfrac{r}{{m - r + 1}} = 2$$ Let us solve the above equation to get the required answer, $$\dfrac{{m - r}}{{r + 1}} = 2 - \dfrac{r}{{m - r + 1}}$$ $$\dfrac{{m - r}}{{r + 1}} = \dfrac{{2m - 2r + 2 - r}}{{m - r + 1}}$$ $${m^2} - mr + m - mr + {r^2} - r = 2mr - 3{r^2} + 2r + 2m - 3r + 2$$ On solving and regrouping the above equation, we get, $${m^2} - m(4r + 1) + 4{r^2} - 2 = 0$$ Hence, (B) is the correct option. **Note:** A combination is a grouping or subset of items. For a combination, $$^n{C_r} = \dfrac{{n!}}{{(n - r)!r!}}$$ Where, factorial n is denoted by n! And defined by n! = n(n-1)(n-2)(n-3)(n-4)…….2.1