gwd11-25

hedgeforfun

GWD11-25.If the sequence x₁, x_2, x_{3, …,} x_n, ?is such that x₁ = 3 and x_n+1 = 2x_n – 1 for n ≥ 1, then x₂₀ – x₁₉ =

A. 2¹⁹

B. 2²⁰

C. 2²¹

D. 2²⁰ - 1

E. 2²¹ - 1

Hot to get answer A, help please.

thinkers

由x_n+1 = 2x_n – 1

x_n+1–x_n= x_n – 1,所以x₂₀ – x₁₉ =x₁₉–1；

又由x_n+1 = 2x_n – 1

x_n+1 = 2（2x_n-1– 1） – 1＝。。。＝2^nx₁–（2^n–1）,

所以x₂₀ – x₁₉ =x₁₉–1＝3*2^18–2^18+1-1＝2*2^18＝2^19

hedgeforfun

many  thanks !!!