A) \[{{-}^{2n}}{{C}_{n-1}}\]
B) \[{{-}^{2n}}{{C}_{n}}\]
C) \[^{2n}{{C}_{n-1}}\]
D) \[^{2n}{{C}_{n}}\]
Correct Answer: D
Solution :
Given expansion can be re-written as \[{{\left( \frac{-1}{x} \right)}^{n}}.{{(1-x)}^{n}}={{(-1)}^{n}}{{x}^{-n}}{{(1-x)}^{2n}}\] Total number of terms will be 2n + 1 which is odd (\[\because 2n\]is playas even) \[\therefore \]Middle term \[=\frac{2n+1+1}{2}=(n+1)th\] Now, \[{{T}_{r+1}}{{=}^{n}}{{C}_{r}}{{(1)}^{r}}{{x}^{n-r}}\] So,\[\frac{^{2n}{{C}_{n}}.{{x}^{2n-n}}}{{{x}^{n}}.{{(-1)}^{n}}}{{=}^{2n}}{{C}_{n}}.{{(-1)}^{n}}\] Middle term is an odd term. So, n + 1 will be odd. So, n will be even. \[\therefore \]Required answer is \[^{2n}{{C}_{n}}.\]
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