Question

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

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

Answer 1

Answer: (C)

$\left(x+\frac{1}{x}\right)+\left(x^{2}+\frac{1}{x^{2}}\right)+\ldots+\left(x^{52}+\frac{1}{x^{52}}\right)$
$=\left(x+x^{2}+\ldots+x^{52}\right)+\left(\frac{1}{x}+\frac{1}{x^{2}}+\ldots+\frac{1}{x^{52}}\right)$
$=x \left(1+x+x^{2}+\ldots+x^{51}\right)+\frac{1}{x^{52}}\left(1+x+\ldots+x^{51}\right)$
$=\left(1+x+x^{2}+\ldots+x^{51}\right)\left(x+\frac{1}{x^{52}}\right)$
$=\frac{1-x^{52}}{1-x}\cdot\frac{x^{53}+1}{x^{52}}$
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 :
$\left(x+x^{2}+\ldots+x^{52}\right)+\left(\frac{1}{x}+\frac{1}{x^{2}}+\frac{1}{x^{3}}+\ldots+\frac{1}{x^{52}}\right)$
Putting $x = \omega$
$\left(\omega+\omega^{2}+\ldots+\omega^{52}\right)+\left(\frac{1}{\omega}+\frac{1}{\omega^{2}}+\ldots+\frac{1}{\omega^{52}}\right)$
$=\omega\left(1+\omega+\ldots+\omega^{51}\right)+\frac{1}{\omega^{52}}\left(1+\omega+\ldots+\omega^{51}\right)$
$=\left(1+\omega+\omega^{2}+\ldots+\omega^{51}\right)\left(\omega+\frac{1}{\omega^{52}}\right)$
$=\left(\frac{1-\omega^{52}}{1-\omega}\right)\left(\omega+\omega\right)^{2}=\left(\frac{1-\omega}{1-\omega}\right)\left(-1\right)=-1$
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.
$x^{52}+\frac{1}{x^{52}}=\omega^{52}+\left(\omega^{2}\right)^{52}=\omega^{52}+\omega^{104}$
$=\omega+\omega^{2}=-1$
$\left(\because1+\omega+\omega^{2}=0\right)$