Math Olympiad | A Nice Algebra Problem | VIJAY Maths

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[@Danieswors]

-1

[@guyhoghton399]

Let _uₙ = xⁿ + ⅟xⁿ_
It is easy to show that _uₙ₊₂ = uₙu₂ - uₙ₋₂_
Given that _u₁ = √3_
_u₂ = (x² + ⅟x²) = (x + ⅟x)² - 2 = u₁² - 2 = 1_
∴ _uₙ₊₂ = uₙ - uₙ₋₂_ ... ①
Since _u₀ = 2_ we can use ① to generate recursively the sequence
_u₀, u₂, u₄, u₆, u₈, u₁₀_ as
_2, 1, -1, -2, -1, 1_
after which the sequence repeats.
∴ _u₁₂ₙ₊ₓ = uₓ_ ( for even _x_ )
∴ *_u₁₀₀ = u₉₆₊₄ = u₄ = -1_*

[@AdwaitaPaul-y2b]

Very good

[@vijaymaths5483]

Thanks

[@RealQinnMalloryu4]

{x+x ➖}=x^2{ 1+1 ➖ }/{x+x ➖ }={x^2+2}/x^2= 2x^2/x^2 =2x^1 (x ➖ 2x+1) .{ x^100+x^100 ➖ }=x^200{1+1 ➖ }/{x^100+ x^100 ➖ } = {x^200+2}/x^200=2x^200/x^200= 2x^1 (x ➖ 2x+1) .