If the sequence x1, x2, x3, ..., xn, ... is such that x1 = 3 and xn+1= 2xn- 1 for n≥ 1, then x^20 - x^19 =
A. 2^19
B. 2^20
C. 2^21
D. 2^20 - 1
E. 2^21 - 1
A. 2^19
B. 2^20
C. 2^21
D. 2^20 - 1
E. 2^21 - 1
x(1) = 3taneja.niks wrote:If the sequence x1, x2, x3, ..., xn, ... is such that x1 = 3 and xn+1= 2xn- 1 for n≥ 1, then x^20 - x^19 =
A. 2^19
B. 2^20
C. 2^21
D. 2^20 - 1
E. 2^21 - 1
x(n+1) = 2x(n) - 1
x(2) = 2x(1) - 1 = 2*3 - 1 = 5
x(3) = 2x(2) - 1 = 2*5 - 1 = 9
x(4) = 2x(3) - 1 = 2*9 - 1 = 17
...
Generalizing, we get that x(n) = 2^n + 1
So, x(20) = 2^20 + 1 and x(19) = 2^19 + 1
Therefore, x^20 - x^19 = 2^20 + 1 - 2^19 - 1 = 2^19(2 - 1) = 2^19
The correct answer is A.
