Question

If $t^2+t+1=0$, then ${(t+\frac{1}{t})}^2+{(t^2+\frac{1}{t^2})}^2+{(t^3+\frac{1}{t^3})}^2....+{(t^{27}+\frac{1}{t^{27}})}^2$ is equal to

• A. $54$
• B. $27$
• C. $18$
• D. $0$

Answer 1

The correct option is A $54$

$t^2+t+1=0$
$\Rightarrow t=\frac{-1+i\sqrt{3}}{2},\frac{-1-i\sqrt{3}}{2}=w,w^2$
Let $t=w\Rightarrow \frac{1}{t}=\frac{1}{w}=w^2$
$\therefore t+\frac{1}{t}=w+w^2=-1$
If $r$ is not divisile by $3$,then
$t^r+\frac{1}{t^r}=w^r+(w^2{)}^r=w^r+w^{2r}=-1$
If $r$ is divisible by $3$,then
$t^r+\frac{1}{t^r}=2$
Hence ${(t+\frac{1}{t})}^2+{(t^2+\frac{1}{t^2})}^2+{(t^3+\frac{1}{t^3})}^2....+{(t^{27}+\frac{1}{t^{27}})}^2$
$=(1+1+1+....18\text{times})+(4+4+4+....9\text{times})=54$