Question

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

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

Answer 1

Answer: (C)

$\left(x+\frac{1}{x}\right)+\left(x^2+\frac{1}{x^2}\right)+\dots +\left(x^{52}+\frac{1}{x^{52}}\right)$
$=\left(x+x^2+\dots +x^{52}\right)+\left(\frac{1}{x}+\frac{1}{x^2}+\dots +\frac{1}{x^{52}}\right)$
$=x\left(1+x+x^2+\dots +x^{51}\right)+\frac{1}{x^{52}}\left(1+x+\dots +x^{51}\right)$
$=\left(1+x+x^2+\dots +x^{51}\right)\left(x+\frac{1}{x^{52}}\right)$
$=\frac{1-x^{52}}{1-x}\cdot \frac{x^{53}+1}{x^{52}}$
on putting $x=\omega$, we get
$=\frac{1-{\omega }^{52}}{1-\omega }\times \frac{\left({\omega }^2+1\right)}{\omega }$
$\left(\because {\omega }^{52}=\omega ,1+{\omega }^2=-\omega \right)$
Alternative Solution :
$\left(x+x^2+\dots +x^{52}\right)+\left(\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}+\dots +\frac{1}{x^{52}}\right)$
Putting $x=\omega$
$\left(\omega +{\omega }^2+\dots +{\omega }^{52}\right)+\left(\frac{1}{\omega }+\frac{1}{{\omega }^2}+\dots +\frac{1}{{\omega }^{52}}\right)$
$=\omega \left(1+\omega +\dots +{\omega }^{51}\right)+\frac{1}{{\omega }^{52}}\left(1+\omega +\dots +{\omega }^{51}\right)$
$=\left(1+\omega +{\omega }^2+\dots +{\omega }^{51}\right)\left(\omega +\frac{1}{{\omega }^{52}}\right)$
$=\left(\frac{1-{\omega }^{52}}{1-\omega }\right){\left(\omega +\omega \right)}^2=\left(\frac{1-\omega }{1-\omega }\right)\left(-1\right)=-1$
Short Cut Method :
$1+x+x^2=0$
$\therefore \left(x-\omega \right)\left(x-{\omega }^2\right)=0$
$\therefore x=\omega ,{\omega }^2=\omega ,\frac{1}{\omega }$
Values of each three sets is 0 i.e
$\left(x+\frac{1}{x}\right)+\left(x^2+\frac{1}{x^2}\right)+\left(x^3+\frac{1}{x^3}\right)=0$
There are $52$ such sets means value up to $51$ sets is 0, we left with the last set i.e.
$x^{52}+\frac{1}{x^{52}}={\omega }^{52}+{\left({\omega }^2\right)}^{52}={\omega }^{52}+{\omega }^{104}$
$=\omega +{\omega }^2=-1$
$\left(\because 1+\omega +{\omega }^2=0\right)$