A) \[\sqrt{3}\]
B) \[\frac{1}{2}\]
C) \[n\]
D) 0
Correct Answer: C
Solution :
As \[1,{{a}_{1}},{{a}_{2}},...,{{a}_{n-1}}\]are nth roots of unity \[\Rightarrow \]\[({{x}^{n}}-1)=(x-1)(x-{{a}_{1}})(x-{{a}_{2}})\] \[...(x-{{a}_{n-1}})\] or \[\frac{{{x}^{n}}-1}{x-1}=(x-{{a}_{1}})(x-{{a}_{2}})...(x-{{a}_{n-1}})\] \[\left\{ as\,\frac{{{x}^{n}}-1}{x-1}={{x}^{n-1}}+{{x}^{n-2}}+...x+1 \right\}\] \[\therefore \] \[{{x}^{n-1}}+{{x}^{n-2}}+...{{x}^{2}}+x+1\] \[=(x-{{a}_{1}})(x-{{a}_{2}})...(x-{{a}_{n-1}})\] Putting \[x=1,\]we get \[1+1+...n\]times \[=(1-{{a}_{1}})(1-{{a}_{2}})...(1-{{a}_{n-1}})\] \[\Rightarrow \]\[(1-{{a}_{1}})(1-{{a}_{2}})...(1-{{a}_{n-1}})=n\]
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