快考了！数列题目求助！麻烦大家乐！

winny莹

GWD14
Q25:
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

leodondon

手排一下就出来了
x1=3
x2=5   x2-x1=2
x3=9   x3-x2=4
x4=17  x4-x3=8

看到规律了吧

winny莹

受教！

chaplain

看来选择A

sdcar2010

x(n+1)= 2x(n) – 1

So x(n+1) -1 = 2*(x(n) -1)

So x(n+1) - x(n) = x(n) -1 =2*(x(n-1) - 1) = (2^2)*(x(n-2) - 1) = (2^(n-1))*(x1-1) = 2^n

winny莹

x(n+1)= 2x(n) – 1

So x(n+1) -1 = 2*(x(n) -1)

So x(n+1) - x(n) = x(n) -1 =2*(x(n-1) - 1) = (2^2)*(x(n-2) - 1) = (2^(n-1))*(x1-1) = 2^n

-- by 会员 sdcar2010 (2011/2/22 22:30:35)

很是不简单！！我就是在这地方被卡主了！！谢过！！