Question

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

• A. 27
• B. 54
• C. 45
• D. 74

Answer 1

The correct option is B 54

Given $t^2+t+1=0$
$\Rightarrow (t-\omega )(t-{\omega }^2)=0\therefore t=\omega ,{\omega }^2$
Now ${(t+\frac{1}{t})}^2+{(t^2+\frac{1}{t^2})}^2+......$ ${(t^{27}+\frac{1}{t^{27}})}^2$ $=\sum _{n=1}^{27}{(t^n+\frac{1}{t^n})}^2$ $=\sum _{n=1}^{27}({\omega }^n+{\omega }^{2n}{)}^2$ $(\because \frac{1}{{\omega }^n}=\frac{{\omega }^{2n}}{{\omega }^{3n}})$ Now $n\ne 3m$ for 18 numbers from 1 to 27 & $n=3n$ for 9 numbers from 1 to 27 $\sum _{n=1}^{27}({\omega }^n+{\omega }^{2n}{)}^2=18(-1{)}^2+9(2{)}^2=54$ $(\because 1+{\omega }^n+{\omega }^{2n}=$ $\{\begin{array}{c}3forn=3m\\ 0forn\ne 3m\end{array}$
$i.e.({\omega }^n+{\omega }^{2n}=\{\begin{array}{c}2forn=3m\\ -1forn\ne 3m\end{array}$