TTGWD9-13

Q13:

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

Answer: A

求思路

x_n+1
                = 2x_n – 1 ---> x_n+1
                – 1= 2(x_n – 1)----->x_n – 1=(x₁ – 1)*2^(n-1)

x₂₀
                – x₁₉ =x₁₉– 1=(3-1)*2^(19-1)=2^19

看不懂楼上的思路，哪个NN来解释一下啊，详细点，我好几年没学数学了，谢谢哦

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

x1=3=2^!+1

x2=5=2^2+1

x3=9=2^3+1

x4=17=2^4+1

:

:

x19=2^19+1

x20=2^20+1

so x₂₀ – x₁₉ = 2^19

now got it?

x1=3=2^!+1

x2=5=2^2+1

x3=9=2^3+1

x4=17=2^4+1

:

:

x19=2^19+1

x20=2^20+1

so x₂₀ – x₁₉ = 2^19

now got it?

谢谢，知道了。

up

二楼一般法,四楼是归纳法.

都是牛人啊!!