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JEE Main · 2025 · 04 Apr 2025 · Shift 2 · Complex Numbers

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

Answer: 11

Solution

With

α = ω

, a non-real cube root of unity,

(ω^{k} + ω^{- k})^{2} = ω^{2 k} + ω^{k} + 2

, which equals

4

when

3 ∣ k

and

1

otherwise. Every three consecutive terms contribute

1 + 1 + 4 = 6

, so for

n = 3 q + 2

the sum is

6 q + 2

. Setting

6 q + 2 = 20

gives

q = 3

and

n = 11

.

JEE Main · 2025 · 04 Apr 2025 · Shift 2 · Integer answer

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