The middle term in the expansion of {left( {1 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The given expression is ${\left( {1 - \frac{1}{x}} \right)^n}{\left( {1 - x} \right)^n}$.This can be rewritten as ${\left( \frac{x-1}{x} \right)^n

Vedclass · Accepted Answer

$ {}^{2n}{C_n}$

Vedclass · Answer

(D) The given expression is ${\left( {1 - \frac{1}{x}} ight)^n}{\left( {1 - x} ight)^n}$.This can be rewritten as ${\left( \frac{x-1}{x} ight)^n} {(1-x)^n} = {\left( \frac{-(1-x)}{x} ight)^n} {(1-x)^n} = {(-1)^n} {x^{-n}} {(1-x)^{2n}}$.The expansion of ${(1-x)^{2n}}$ has $2n+1$ terms.The middle term is the ${\left( \frac{2n+1+1}{2} ight)}^{	ext{th}} = {(n+1)}^{	ext{th}}$ term.The general term $T_{r+1}$ in the expansion of ${(1-x)^{2n}}$ is given by ${}^{2n}C_r {(-x)^r} = {}^{2n}C_r {(-1)^r} {x^r}$.For the $(n+1)^{	ext{th}}$ term,we set $r=n$,which gives ${}^{2n}C_n {(-1)^n} {x^n}$.Substituting this back into the expression: ${(-1)^n} {x^{-n}} \cdot {}^{2n}C_n {(-1)^n} {x^n} = {(-1)^{2n}} {}^{2n}C_n = {}^{2n}C_n$ (since $2n$ is even,${(-1)^{2n}} = 1$).

The middle term in the expansion of ${\left( {1 - \frac{1}{x}} \right)^n}{\left( {1 - x} \right)^n}$ in powers of $x$ is

Similar Questions

If the coefficients of the middle terms in the binomial expansions of $(1+\alpha x)^{26}$ and $(1-\alpha x)^{28}$, where $\alpha \neq 0$, are equal, then the value of $\alpha$ is:

If $\left(\frac{3^{6}}{4^{4}}\right) k$ is the term independent of $x$ in the binomial expansion of $\left(\frac{x}{4}-\frac{12}{x^{2}}\right)^{12}$,then $k$ is equal to ...... .

If the coefficients of $x^7$ in $\left(ax^2+\frac{1}{2bx}\right)^{11}$ and $x^{-7}$ in $\left(ax-\frac{1}{3bx^2}\right)^{11}$ are equal,then

If the coefficients of $(2 \alpha+4)$-th and $(\alpha-2)$-th terms in the expansion of $(1+x)^{2018}$ are equal,then $\alpha=$

The two middle terms in the expansion of ${\left( {x - \frac{1}{x}} \right)^{11}}$ are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module