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gmat_killer
- Newbie | Next Rank: 10 Posts
- Posts: 6
- Joined: Sat Jun 06, 2009 4:06 pm
- Location: Toronto
Anyone know the answer to this?
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
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
Pain is my motivation, failure is my strength.


















