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Sequence Question

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Newbie | Next Rank: 10 Posts
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by gmat_killer » Sat Jun 06, 2009 4:24 pm
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
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by DanaJ » Sun Jun 07, 2009 5:04 am
See picture...
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by pathaniaus » Sun Jun 07, 2009 6:03 am
Hey DanaJ,

So I looked at the picture... and I still do not understand. Could you explain in detail please?

Thanks!!!
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by kris610 » Sun Jun 07, 2009 6:27 am
Write the down the sequence for the first few values and you get:

3 5 9 17.

Now, in a sequence problem, you will always see a pattern, and here the pattern is, for a given position n, the value is (2^n)+1. For example, for the first position, the value is (2^1)+1 = 3.

So, for position 20 the value will be (2^20)+1 and for 19, (2^19)+1.

Subtract, you get 2^20 - 2^19. Simplify, you get 2^19.
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