∵α−β=5a1+a2+an=∑ai=∑αi−∑βiα−β=α(1−αn)(1−α)−β(1−βn)(1−β)α−β=(α+1)(1−αn)−(β+1)(1−βn)(1−α)(1−β)(α−β)=α2−αn+2−β2+βn+2(1−α)(1−β)(α−β)=5 (4)∑n=1∞bn10n=∑(α10)n+∑(β10)n=α101−α10+β101−β10=α10−α+β10−β=10(α+β)−2;αβ100−1;0(α+β)+αβ=10+289=1289Hence, Options A, B, and C are correct.

" /> ∵α−β=5a1+a2+an=∑ai=∑αi−∑βiα−β=α(1−αn)(1−α)−β(1−βn)(1−β)α−β=(α+1)(1−αn)−(β+1)(1−βn)(1−α)(1−β)(α−β)=α2−αn+2−β2+βn+2(1−α)(1−β)(α−β)=5 (4)∑n=1∞bn10n=∑(α10)n+∑(β10)n=α101−α10+β101−β10=α10−α+β10−β=10(α+β)−2;αβ100−1;0(α+β)+αβ=10+289=1289Hence, Options A, B, and C are correct.

" />
devarshi-dt-logo

Question:

Let α and β be the roots of x² - x - 1 = 0 with α > β. For all positive integers n, define aₙ = αⁿ - βⁿ, n ≥ 1. b₁ = 1 and bₙ = aₙ₋₁ + aₙ₊₁, n ≥ 2. Which of the following options is/are correct?

bₙ = αⁿ + βⁿ for all n ≥ 1

∑ₙ₌₁¹⁰ aₙ10ⁿ = 1089

∑ₙ₌₁¹⁰ bₙ10ⁿ = 889

a₁ + a₂ + ... + aₙ = aₙ₊₂ for all n ≥ 1

Solution:

Correct option is C.a1+a2+an=an+2for alln≥1(1)∑n=1∞an10n=∑αn−βn(α−β)10n==1α−β(α101−α10−β101−β10)1α−β(α10−α−=1α−β (10(α−β)−αβ+αβ)100−1;0(α+β)+αβ=1089(2)bn=an+1+an−1;=αn+1−βn+1α−β+αn−1;−βn−1;α−β=αn−1;(α2+1)α=αn−1;(α+2)−βn−1;(β+2)α−β=αn−1;(5+52)−βn−1;(5−52)α−β5αn−1;(5+12)−5βn−1;(5−1;2)α−β=5(αn+βn)α−β=αn+βn∵α−β=5a1+a2+an=∑ai=∑αi−∑βiα−β=α(1−αn)(1−α)−β(1−βn)(1−β)α−β=(α+1)(1−αn)−(β+1)(1−βn)(1−α)(1−β)(α−β)=α2−αn+2−β2+βn+2(1−α)(1−β)(α−β)=5 (4)∑n=1∞bn10n=∑(α10)n+∑(β10)n=α101−α10+β101−β10=α10−α+β10−β=10(α+β)−2;αβ100−1;0(α+β)+αβ=10+289=1289Hence, Options A, B, and C are correct.