A) \[8\]
B) \[10\]
C) \[12\]
D) \[14\]
Correct Answer: A
Solution :
We have,\[{{x}^{2}}-x+1=0\] \[\Rightarrow \] \[x=\frac{1\pm \sqrt{1-4}}{2}\] \[\Rightarrow \] \[x=\frac{1\pm \sqrt{3}i}{2}\] \[\Rightarrow \] \[x=-\omega ,\,\,-{{\omega }^{2}}\] Now, \[\sum\limits_{n=1}^{5}{{{\left( {{x}^{n}}+\frac{1}{{{x}^{n}}} \right)}^{2}}}\] \[=\sum\limits_{n=1}^{s}{\left( {{x}^{2n}}+\frac{1}{{{x}^{2n}}}+2 \right)}\] \[=\left( {{x}^{2}}+\frac{1}{{{x}^{2}}}+2 \right)+\left( {{x}^{4}}+\frac{1}{{{x}^{4}}}+2 \right)\] \[+\left( {{x}^{6}}+\frac{1}{{{x}^{6}}}+2 \right)+\left( {{x}^{8}}+\frac{1}{{{x}^{8}}}+2 \right)\] \[+\left( {{x}^{10}}+\frac{1}{{{x}^{10}}}+2 \right)\] \[=({{x}^{2}}+{{x}^{4}}+{{x}^{6}}+{{x}^{8}}+{{x}^{10}})\] \[+\left( \frac{1}{{{x}^{2}}}+\frac{1}{{{x}^{4}}}+\frac{1}{{{x}^{6}}}+\frac{1}{{{x}^{8}}}+\frac{1}{{{x}^{10}}} \right)+10\] \[=({{\omega }^{2}}+{{\omega }^{4}}+{{\omega }^{6}}+{{\omega }^{8}}+{{\omega }^{10}})\] \[+\left( \frac{1}{{{\omega }^{2}}}+\frac{1}{{{\omega }^{4}}}+\frac{1}{{{\omega }^{6}}}+\frac{1}{{{\omega }^{8}}}+\frac{1}{{{\omega }^{10}}} \right)+10\]\[(\because \,\,x=\omega )\] \[={{\omega }^{2}}+\omega +1+{{\omega }^{2}}+\omega +\frac{1}{{{\omega }^{2}}}+\frac{1}{\omega }+1+\frac{1}{{{\omega }^{2}}}+\frac{1}{\omega }\] \[+10\left[ \begin{matrix} \because \,\,{{\omega }^{3}}=1\,\,and \\ {{\omega }^{2}}+\omega +1=0 \\ \end{matrix} \right]\] \[=({{\omega }^{2}}+\omega +1)+{{\omega }^{2}}+\omega +\omega +{{\omega }^{2}}+1+\omega +{{\omega }^{2}}\]\[+10\] \[=0+2(\omega +{{\omega }^{2}})+({{\omega }^{2}}+\omega +1)+10\] \[=0+2(-1)+0+10\] \[=-2+10=8\]
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