请教GWD 11-25

qwrersaa

Q25:

If the sequence x1, x2, x3, …, xn, … is such that x1 = 3 and xn+1 = 2xn – 1 for n ≥ 1, then x20 – x19 =

A. 219

B. 220

C. 221

D. 220 - 1

E. 221 - 1

请问如何解答,我先把X20=2X19-1

然后就是2X19-1-X19= X19-1

然后难道要一个一个的算出么,谢谢

mary1

答案是不是选A呀?

因为根据题目，X1=3, X2=2*1-1=5,X3=2*5-1=9.

X2-X1=5-3=2^1

X3-X2=9-5=2^2

.....

X20-X19=2^19

mikejia

X20-X19=X20-X19

=2^0 X19-2^0

=2^1 X18-2^1

=2^2 X17-2^2

......

=2^17 X2-2^17

=2^18 X1-2^18

=2^18(3-1)

=2^19

我算的复杂点哦

qwrersaa

谢谢两位的慷慨解答，我会了