The middle term in the expansion of&nbsp | vedclass

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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

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