我得2^19 - 1 可偏偏没这个选项

supermanlee

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

A.    2^19
B.    2^20
C.    2^21
D.    2^20 - 1
E.    2^21 - 1

卮言浅夏

x1 -1 = 2; then 2^(n-1)*2 = 2^n

supermanlee

这么说X1=3 这个条件不成白给的了吗

supermanlee

没看懂楼上的解释……

卮言浅夏

xn+1 = 2xn – 1

So xn+1 - xn = xn -1 = (2xn-1 -1) -1 = 2xn-1 - 2 = 2(2xn-2 - 1)  -2 = (2^2)*[xn-2 -1] = ..... =(2^(n-1))*[x1 -1]
=2^n


AA