JEE Main 2025 (Online) 4th April Evening ShiftQuestion 1
If $\alpha$ is a root of the equation $x^{2}+x+1=0$ and $\sum_{k=1}^{n}\left(\alpha^{k}+\frac{1}{\alpha^{k}}\right)^{2}=20$, then $n$ is equal to $\_\_\_\_\_\_$ .
Complex Numbers
Answer$11$
Solution
$\alpha=\omega$, a primitive cube root of unity, so $\alpha^k+\alpha^{-k}=2$ if $3\mid k$, else $-1$. $(\alpha^k+\alpha^{-k})^2=4$ if $3\mid k$, else $1$. Each block of 3 consecutive $k$ contributes $1+1+4=6$. After $n=9$ the sum is $18$; $k=10$ adds $1$ (19), $k=11$ adds $1$ (20). Answer: $n=11$.