JEE MAIN - Mathematics (2025 - 4th April Evening Shift - No. 24)

If $\alpha$ is a root of the equation $x^2+x+1=0$ and $\sum_\limits{\mathrm{k}=1}^{\mathrm{n}}\left(\alpha^{\mathrm{k}}+\frac{1}{\alpha^{\mathrm{k}}}\right)^2=20$, then n is equal to _________.
Answer
11

Explanation

$$\begin{aligned} &\alpha \text { is root of equation } 1+x+x^2=0, \alpha=\omega \text { or } \omega^2\\ &\begin{aligned} & \left(\alpha^k+\frac{1}{\alpha^k}\right)^2=\alpha^{2 k}+\frac{1}{\alpha^{2 k}}+2=\omega^k+\frac{1}{\omega^k}+2 \\ & \Rightarrow \quad \omega^k+\frac{1}{\omega^k}+2=\left\{\begin{array}{l} 4,3 \text { divides } k \\ 1,3 \text { does not divide } k \end{array}\right. \\ & \therefore \quad \sum_{k=1}^n\left(\alpha^k+\frac{1}{\alpha^k}\right)^2=20 \\ & \Rightarrow \quad(1+1+4)+(1+1+4)+(1+1+4)+(1+1) \\ & =20 \\ & \Rightarrow \quad n=11 \end{aligned} \end{aligned}$$

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