If is a root of the equation and , then n is equal to __________. [2025]

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Solution

Q.
If $α$ is a root of the equation $x^{2} + x + 1 = 0$ and $\sum_{k = 1}^{n} {(α^{k} + \frac{1}{α^{k}})}^{2} = 20$ , then n is equal to __________. [2025]

Ans.
(11)

We have, $α$ is root of equation $x^{2} + x + 1 = 0$ .

Let $α = ω$ .

Then, $\sum_{k = 1}^{n} {(ω^{k} + \frac{1}{ω^{k}})}^{2} = \sum_{k = 1}^{n} (ω^{2 k} + \frac{1}{ω^{2 k}} + 2)$

$= \sum_{k = 1}^{n} (ω^{2 k} + ω^{k} + 2) (∵ ω^{3 k} = 1)$

But $\sum_{k = 1}^{n} (ω^{2 k} + ω^{k} + 2) = 20$

$\Rightarrow (ω^{2} + ω^{4} + . . . + ω^{2 n}) + (ω + ω^{2} + . . . + ω^{n}) + 2 n = 20$ .

If n = 3m, m $\in$ $I$

Then, 0 + 0 + 2n = 20

$\Rightarrow n = 10$ (not satisfy as m is integer).

If n = 3m + 1, then $ω^{2} + ω + 2 n = 20$ .

$\Rightarrow n = \frac{21}{2}$ (not possible)

If n = 3m + 2, then $(ω^{2} + ω^{4}) + (ω + ω^{2}) + 2 n = 20$ .

$\Rightarrow 2 (ω^{2} + ω + n) = 20$ .

$\Rightarrow n = 10 + 1 = 11$ .

$∴$ For n = 11 satisfies n = 3m + 2.

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