The middle term in the expansion of&nbsp | vedclass

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

Question

Given expansion can be re-written as

$\left(\frac{x-1}{x}\right)^{n} \cdot(1-x)^{n}=(-1)^{n} x^{-n}(1-x)^{2 n}$

Total number of terms will be $2

Vedclass · Accepted Answer

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

Vedclass · Answer

Given expansion can be re-written as

$\left(\frac{x-1}{x}ight)^{n} \cdot(1-x)^{n}=(-1)^{n} x^{-n}(1-x)^{2 n}$

Total number of terms will be $2n+1$ which is odd

$(\because 2 n 	ext { is always even })$

$	ext { Middle term }=\frac{2 n+1+1}{2}=(n+1) 	ext { th }$

Now, $T_{r+1}=^{n} C_{r}(1)^{r} x^{n-r}$

So, $\frac{2 n_{C_{n}-x^{2 n-n}}}{x^{n} \cdot(-1)^{n}}=^{2 n} C_{n} \cdot(-1)^{n}$

Middle term is an odd term So, $n+1$ will be odd.

So, $n$ will be even.

$	herefore $ Required answer is $^{2 n}{C}_{n}$

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

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

The middle term in the expansion of ${(1 + x)^{2n}}$ is

If sum of the coefficient of the first, second and third terms of the expansion of ${\left( {{x^2} + \frac{1}{x}} \right)^m}$ is $46$, then the coefficient of the term that doesnot contain $x$ is :-

The second, third and fourth terms in the binomial expansion $(x+a)^n$ are $240,720$ and $1080,$ respectively. Find $x, a$ and $n$

The coefficient of $x^4$ in ${\left[ {\frac{x}{2}\,\, - \,\,\frac{3}{{{x^2}}}} \right]^{10}}$ is :