Question

lf $x^2+x+1=0$, then the value of
$(x+\frac{1}{x})+(x^2+\frac{1}{x^2})+\dots +(x^{52}+\frac{1}{x^{52}})$ equals

• A. $1$
• B. $0$
• C. $-1$
• D. $-52$

Answer 1

Answer: (C)

$-1$

$\omega$ and ${\omega }^2$ are the roots of the given equation.

We have,

$x+\frac{1}{x}=\omega +{\omega }^2=-1$

$x^2+\frac{1}{x^2}={\omega }^2+\omega =-1$

$x^3+\frac{1}{x^3}=2$

This pattern will repeat after an interval of three terms. Hence, the sum of the given series is the last term.

Sum $=-1$