Question

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

Answer 1

Hint:Here we are going to find the coefficient of \[{x^n}\] 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,
\[{(x + y)^n} = \sum\limits_{k = 0}^n {^n{C_k}{x^{n - k}}{y^k}} = {x^n}{ + ^n}{C_1}{x^{n - 1}}y{ + ^n}{C_2}{x^{n - 2}}{y^2} + ......\]

Complete step-by-step answer:
We have to find out the coefficient of \[{x^n}\] in expansion of\[(1 + x){(1 - x)^n}\] .
From Binomial theorem we have,
\[{(x + y)^n} = \sum\limits_{k = 0}^n {^n{C_k}{x^{n - k}}{y^k}} = {x^n}{ + ^n}{C_1}{x^{n - 1}}y{ + ^n}{C_2}{x^{n - 2}}{y^2} + ......\]
Let us put x=1, y=-x in the above equation and apply binomial theorem to get the expansion of \[{(1 - x)^n}\]
\[{(1 - x)^n} = \sum\limits_{k = 0}^n {^n{C_k}{{(1)}^{n - k}}{{( - x)}^k}} \]
And on expanding the summation in the above equation we get,\[{(1 - x)^n} = 1{ + ^n}{C_1}{(1)^{n - 1}}( - x){ + ^n}{C_2}{(1)^{n - 2}}{( - x)^2} + ......{ + ^n}{C_{n - 2}}{(1)^{n - (n - 2)}}{( - x)^{n - 2}}{ + ^n}{C_{n - 1}}{(1)^{n - (n - 1)}}{( - x)^{n - 1}} + {( - x)^n}\]Thus we can see from expansion of \[(1 + x){(1 - x)^n}\]
\[(1 + x){(1 - x)^n} = 1.{(1 - x)^n} + x{(1 - x)^n}\]
Let us use the binomial expansion so that we get,
\[
  (1 + x){(1 - x)^n} \\
   = 1.\{ 1{ + ^n}{C_1}{(1)^{n - 1}}( - x){ + ^n}{C_2}{(1)^{n - 2}}{( - x)^2} + ......{ + ^n}{C_{n - 2}}{(1)^{n - (n - 2)}}1.{( - x)^{n - 2}}{ + ^n}{C_{n - 1}}{(1)^{n - (n - 1)}}1.{( - x)^{n - 1}} + {( - x)^n}\} \\
   + x\{ 1{ + ^n}{C_1}{(1)^{n - 1}}( - x){ + ^n}{C_2}{(1)^{n - 2}}{( - x)^2} + ......{ + ^n}{C_{n - 2}}{(1)^{n - (n - 2)}}1.{( - x)^{n - 2}}{ + ^n}{C_{n - 1}}{(1)^{n - (n - 1)}}1.{( - x)^{n - 1}} + {( - x)^n}\} \\
\]
The coefficient of \[{x^n}\] in expansion of \[(1 + x){(1 - x)^n}\]is
\[{( - 1)^n}{ + ^n}{C_{n - 1}}{(1)^{n - (n - 1)}}1.{( - 1)^{n - 1}}\]
Let us solve using the combination formula we get,
\[{( - 1)^n}{ + ^n}{C_{n - 1}}{(1)^{n - (n - 1)}}1.{( - 1)^{n - 1}} = {( - 1)^n} + \dfrac{{n!}}{{(n - 1)!(n - n + 1)!}}1.{( - 1)^{n - 1}}\]
On further simplifications using factorial we get
\[{( - 1)^n} + \dfrac{{n(n - 1)!}}{{(n - 1)!.1}}1.{( - 1)^{n - 1}} = {( - 1)^{n - 1}}( - 1 + n)\]
We further modify it to find the answer,
\[{( - 1)^n}{ + ^n}{C_{n - 1}}{(1)^{n - (n - 1)}}1.{( - 1)^{n - 1}} = {( - 1)^n}(1 - n)\]
Hence, (B) is the correct option.

We have to find out the coefficient of \[{r^{th}},{(r + 1)^{th}}and{(r + 2)^{th}}\] terms in the binomial expansion\[{(1 + y)^m}\].
From Binomial theorem we have,
\[{(x + y)^n} = \sum\limits_{k = 0}^n {^n{C_k}{x^{n - k}}{y^k}} = {x^n}{ + ^n}{C_1}{x^{n - 1}}y{ + ^n}{C_2}{x^{n - 2}}{y^2} + ......\]
Let us put x=1, n=m and apply binomial theorem to get the expansion of \[{(1 + y)^m}\]
\[\begin{gathered}
  {(1 + y)^m} = \sum\limits_{k = 0}^m {^m{C_k}{1^{m - k}}{y^k}} \\
   = 1{ + ^m}{C_1}{1^{m - 1}}y{ + ^m}{C_2}{1^{m - 2}}{y^2} + {......^m}{C_{r - 1}}{1^{m - (r - 1)}}{y^{r - 1}}{ + ^m}{C_r}{1^{m - r}}{y^r}{ + ^m}{C_{r + 1}}{1^{m - (r + 1)}}{y^{r + 1}} + .... \\
\end{gathered} \]
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