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7.Binomial Theorem
hard
${\left( {1 - \frac{1}{x}} \right)^n}\left( {1 - {x}} \right)^n$ ના વિસ્તરણમાં મધ્યમ પદ મેળવો.
A
$ - {}^{2n}{C_{n - 1}}$
B
$ - {}^{2n}{C_n}$
C
$ {}^{2n}{C_{n - 1}}$
D
$ {}^{2n}{C_n}$
(AIEEE-2012)
Solution
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 $2n+1$ which is odd
$(\because 2 n \text { is always even })$
$\text { Middle term }=\frac{2 n+1+1}{2}=(n+1) \text { th }$
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.
$\therefore $ Required answer is $^{2 n}{C}_{n}$
Standard 11
Mathematics
