If α is a root of the equation x^2+x+1=0 and ∑_ k=1 ^ n (α^ k + 1 α^ k )^2=20 , then n is equal to

Correct answer: 0 aligned & α=ω & (ω^k+(1)/(ω^k))^2=ω^ 2 k + 1 ω^ 2 k +2 & =ω^ 2 k +ω^k+2 ω^ 3 k =1 aligned aligned & ∑_ k=1 ^ n (ω^ 2 k +ω^ k +2)=20 & (ω^2+ω^4+ω^6+ +ω^ 2 n )+(ω+ω^2+ω^3+ +. & .ω^ n )+2 n=20 aligned Now if n=3 m, m ∈ I Then 0+0+2 n=20 n=10 (not satisfy) if n=3 m+1 , then aligned & ω^2+ω+2 n=20 & -1+2 n=20 n=(21)/(2)( not possible ) aligned if n=3 m+2 , aligned & (ω^8+ω^ 10 )+(ω^4+ω^5)+2 n=20 & (ω^2+ω)+(ω+ω^2)+2 n=20 & 2 n=22 & n=11 satisfy n=3 m+2 & n=11 aligned

JEE Main 20254 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 4 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Complex Number

JEE Main 2025 · 4 April · Evening Shift · Mathematics · Complex Number

If α is a root of the equation x^2+x+1=0 and ∑_ k=1 ^ n (α^ k + 1...

\alpha

is a root of the equation

x^2+x+1=0

and

\sum_{\mathrm{k}=1}^{\mathrm{n}}\left(\alpha^{\mathrm{k}}+\frac{1}{\alpha^{\mathrm{k}}}\right)^2=20

, then n is equal to

Correct answer

Step-by-step solution

\begin{aligned} & \alpha=\omega \\ & \therefore\left(\omega^k+\frac{1}{\omega^k}\right)^2=\omega^{2 k}+\frac{1}{\omega^{2 k}}+2 \\ & =\omega^{2 k}+\omega^k+2 \quad \because \omega^{3 k}=1\end{aligned}

\begin{aligned}& \therefore \sum_{\mathrm{k}=1}^{\mathrm{n}}\left(\omega^{2 \mathrm{k}}+\omega^{\mathrm{k}}+2\right)=20 \\ & \Rightarrow\left(\omega^2+\omega^4+\omega^6+\ldots+\omega^{2 \mathrm{n}}\right)+\left(\omega+\omega^2+\omega^3+\ldots+\right. \\ & \left.\omega^{\mathrm{n}}\right)+2 \mathrm{n}=20\end{aligned}

Now if

n=3 m, \quad m \in I

Then

0+0+2 \mathrm{n}=20 \Rightarrow \mathrm{n}=10

(not satisfy)

if n=3 m+1

, then

\begin{aligned}& \omega^2+\omega+2 \mathrm{n}=20 \\ & -1+2 \mathrm{n}=20 \Rightarrow \mathrm{n}=\frac{21}{2}(\text{ not possible })\end{aligned}

if

n=3 m+2

,

\begin{aligned}& \left(\omega^8+\omega^{10}\right)+\left(\omega^4+\omega^5\right)+2 \mathrm{n}=20 \\ & \Rightarrow\left(\omega^2+\omega\right)+\left(\omega+\omega^2\right)+2 \mathrm{n}=20 \\ & 2 \mathrm{n}=22 \\ & \mathrm{n}=11 \text{ satisfy } \mathrm{n}=3 \mathrm{ m}+2 \\ & \therefore \mathrm{n}=11\end{aligned}

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