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
A. 2^19
B. 2^20
C. 2^21
D. 2^20 - 1
E. 2^21 - 1
Given Data:
X1=3
Xn+1=2Xn-1------(1)
n=1 or n>1
if n=1 then (1) becomes
X2=6
X3=12
X4=24......So on
This is in Geometric progression so
Tn=ar^n-1
where r is the common ratio
a is the first term
X20 is calculated by the formula
X20=3*2^20-1 here a=3 (ieX1=3)-----2
similarly
X19=3*2^19-1-------3
2-3 gives
3*2^19-3*2^18====>3(2^19-2^18)==>3*2^18 on further simplify.....
I think either question or answer choices are wrong
