Question

If $t^2+t+1$ = 0 then prove that the value of
${(t+\frac{1}{t})}^2+{(t^2+\frac{1}{t^2})}^2+....{(t^{27}+\frac{1}{t^{27}})}^2=54$

Answer 1

Here x =$\omega$,${\omega }^2$.

As in last question for x = $\omega$

$\underset{n=27}{\sum ^5}{(t^n+\frac{1}{t^n})}^2=\underset{n=27}{\sum ^5}({\omega }^n+{\omega }^{2n}{)}^2$

Now $n\ne 3p$ for 18 numbers from 1 to 27 and n = 3p for 9 numbers from 1 to 27

$\therefore \underset{n=27}{\sum ^5}=18(-1{)}^2+9(2{)}^2=18+36=54$

The same result will be true for x = ${\omega }^2$