JEE Main 2026 — Jan 21, Shift 1 — Mathematics solution

JEE Main2026Jan 21, Shift 1Mathematics

If $x^{2} + x + 1 = 0$ , then the value of $(x + \frac{1}{x})^{4} + (x^{2} + \frac{1}{x ^{2}})^{4} + (x^{3} + \frac{1}{x ^{3}})^{4} + \dots + (x^{25} + \frac{1}{x ^{25}})^{4}$ is :

A
128
B
162
C
175
D
145

Solution

$x$ is a primitive cube root of unity ( $x = ω$ ), so $1/ x = x^{2} = \overset{ω}{ˉ}$ . Hence $x^{n} + x^{- n} = ω^{n} + ω^{- n}$ , which depends only on $n mod 3$ :

If $3 ∣ n$ : $ω^{n} + ω^{- n} = 1 + 1 = 2$ , so the term contributes $2^{4} = 16$ .

Otherwise: $ω^{n} + ω^{- n} = - 1$ , so the term contributes $(- 1)^{4} = 1$ .

Among $n = 1, 2, \dots, 25$ there are $⌊ 25/3 ⌋ = 8$ multiples of 3, and the other $25 - 8 = 17$ values contribute 1 each.

Total $= 17 \cdot 1 + 8 \cdot 16 = 17 + 128 = 145$ .

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