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Mathematics · Sequence and Series

JEE Main 2025 — 4 April, Evening Shift — Question 38

If α\alpha is a root of the equation x2+x+1=0x^{2}+x+1=0 and ∑k=1n(αk+1ak)2=20\sum_{k=1}^{n}\left(\alpha^{k}+\frac{1}{a^{k}}\right)^{2}=20, then nn is equal to ____\_\_\_\_ .

Answer: 11

Numerical answer — enter this value.

Step-by-step solution

α\alpha is root of equation 1+x+x2=0,α=ω1+x+x^{2}=0, \alpha=\omega or ω2\omega^{2}

(αk+1αk)2=α2k+1α2k+2=ωk+1ωk+2⇒ωk+1ωk+2={4,3 divides k1,3 does not divide k∴∑k=1n(αk+1αk)2=20⇒(1+1+4)+(1+1+4)+(1+1+4)+(1+1)=20⇒n=11\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 n=11 \end{aligned}

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Exam
JEE Main 2025
Paper
4 April, Evening Shift
Subject
Mathematics
Chapter
Sequence and Series
Topic
Introduction to Sequence and Series

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