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 $4^{\text{th}}$ term in the expansion of $\left(\frac{x}{2}-\frac{2y}{3}\right)^6$ is $-20$,then $xy=$

If the sum of the coefficients of $x^r$ $(r=0, 1, 2, \ldots, 2n)$ in the expansion of $(1+3x-2x^2)^n$ is $128$,then $\sum_{r=1}^{2n} r \frac{^{2n}C_r}{^{2n}C_{r-1}} = $

The sum of the rational terms in the binomial expansion of $(\sqrt{2} + 3^{1/5})^{10}$ is

Given that the $4^{th}$ term in the expansion of $(2 + \frac{3}{8}x)^{10}$ has the maximum numerical value,the range of values of $x$ for which this will be true is given by:

In the binomial $(2^{1/3} + 3^{-1/3})^n$,if the ratio of the seventh term from the beginning of the expansion to the seventh term from its end is $1/6$,then $n =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module