Question

If $x^2+x+1=0$, then the value of $(x+\frac{1}{x})+(x^2+\frac{1}{x^2})+....+(x^{52}+\frac{1}{x^{52}})$ is equal to

• A. $1$
• B. $0$
• C. $-1$
• D. none of these

Answer 1

Answer: (C)

$-1$

Given $x^2+x+1=0$
$\Rightarrow x=\omega ,{\omega }^2$
Now, $(x+\frac{1}{x})+(x^{{}_2}+\frac{1}{x^2})+...+(x^{52}+\frac{1}{x^{52}})$
$=(x+x^2+...+x^{52})+(\frac{1}{x}+\frac{1}{x^2}+...\frac{1}{x^{52}})$
$=x(1+x+x^2+...+x^{52})+\frac{1}{x^{52}}(1+x+...+x^{52})$ $=(1+x+x^2+...+x^{52})(x+\frac{1}{x^{52}})$
$=\frac{1-x^{52}}{1-x}\cdot \frac{x^{53}+1}{x^{52}}$
Put $x=\omega$ we get
$=\frac{1-{\omega }^{52}}{1-\omega }\times \frac{({\omega }^2+1)}{\omega }({\omega }^{52}=\omega ,1+{\omega }^2=-\omega )$
$=-1$