Problem Solving

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

[spoiler]OA-A
IMO-C[/spoiler]

I doubt the OA

x1 = 3
x2 = 2(3) - 1 = 5
x3 = 2(5) - 1 = 9
x4 = 2(9) - 1 = 17
x5 = 2(17) - 1 = 33

Now we can write the sequence as

x1 = 2^1 + 1
x2 = 2^2 + 1
x3 = 2^3 + 1
x4 = 2^4 + 1
x5 = 2^5 + 1
.
.
.
.
x19 = 2^19 + 1
x20 = 2^20 + 1

x20 - x19 = 2^20 + 1 - (2^19 + 1)
= 2^20 + 1 - 2^19 - 1
= 2^20 - 2^19
= 2^19 (2 - 1)
= 2^19

Choose A

From the info given, we can observe that
x1= 3, x2=4, x3=8, x4=17...
thus, x2-x1=2
x3-x2=4 i.e 2^2
x4-x3=8 i.e 2^3

thus we can see that x(n+1)-xn=2^n

so X20-X19= 2^19.

Here n+1 is 20 and n is 19.