A) \[{{\left[ \frac{1-x}{1-2x} \right]}^{n}}\]
B) \[{{(1-x)}^{n}}\]
C) \[{{\left[ \frac{1-2x}{1-x} \right]}^{n}}\]
D) \[{{\left( \frac{1}{1-x} \right)}^{n}}\]
Correct Answer: A
Solution :
[a] Given that \[1+n\left| \frac{x}{1-x} \right|+\frac{n(n+1)}{2!}{{\left| \frac{x}{1-x} \right|}^{2}}\] \[+....\infty \] is expansion of \[{{\left| 1-\frac{x}{1-x} \right|}^{-n}}\]. So, it is \[={{\left| 1-\frac{x}{1-x} \right|}^{-n}}\] \[={{\left| \frac{1-x-x}{1-x} \right|}^{-n}}={{\left| \frac{1-x}{1-2x} \right|}^{n}}\]You need to login to perform this action.
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