A) \[n\]
B) \[n-1\]
C) \[-1\]
D) \[1\]
Correct Answer: A
Solution :
Since,\[1,{{\alpha }_{1}},{{\alpha }_{2}},....,{{\alpha }_{n-1}}\]are the roots of unity. \[\therefore \]\[(x-1)(x-{{\alpha }_{1}})(x-{{\alpha }_{2}}).....(x-{{\alpha }_{n-1}})={{x}^{n}}-1\] \[\Rightarrow \]\[(x-{{\alpha }_{1}})(x-{{\alpha }_{2}})....(x-{{\alpha }_{n-1}})=\frac{{{x}^{n}}-1}{x-1}\] \[\Rightarrow \]\[(x-{{\alpha }_{1}})(x-{{\alpha }_{2}})....(x-{{\alpha }_{n-1}})\] \[={{x}^{n-1}}+{{x}^{n-2}}+.....+x+1\] Put\[x=1\]on both sides, \[(1-{{\alpha }_{1}})(1-{{\alpha }_{2}})....(1-{{\alpha }_{n-1}})\] \[={{1}^{n-1}}+{{1}^{n-2}}+....+1+1\] \[=n\]
You need to login to perform this action.
You will be redirected in
3 sec
