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