A) \[\sqrt{3}\]
B) \[\frac{1}{2}\]
C) \[n\]
D) \[0\]
Correct Answer: C
Solution :
Since,\[1,\,\,{{a}_{1}},\,\,{{a}_{2}},\,\,....,\,\,{{a}_{n-1}}\]are the nth root of unity. \[\therefore \] \[{{x}^{n}}-1=(x-1)(x-{{a}_{1}})...(x-{{a}_{n-1}})\] \[\Rightarrow \] \[\frac{{{x}^{n}}-1}{x-1}=(x-{{a}_{1}})(x-{{a}_{2}})...(x-{{a}_{n-1}})\] \[\therefore \] \[{{x}^{n-1}}+{{x}^{n-2}}+...+{{x}^{2}}+x+1\] \[=(x-{{a}_{1}})(x-{{a}_{2}})...(x-{{a}_{n-1}})\] Put\[x=1\], we get \[(1-{{a}_{1}})(1-{{a}_{2}})...(1-{{a}_{n-1}})=1+1+...+n\]times \[=n\]
You need to login to perform this action.
You will be redirected in
3 sec
