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If the sequence x{_1} , x{_2} , x{_3} , ...., x{_n} , ....is [#permalink]
17 Jul 2008, 02:03

If the sequence \(x{_1}\), \(x{_2}\), \(x{_3}\), ...., \(x{_n}\), ....is such that \(x{_1}\) = 3 and \(x{_n_+_1}\) = \(2x{_n} - 1\) for n >= 1, then \(x{_2_0}\) - \(x{_1_9}\) =

A. \(2^{19}\) B. \(2^{20}\) C. \(2^{21}\) D. \(2^{20} -1\) E. \(2^{21} -1\) _________________

Re: Math Set 1: Q25 [#permalink]
17 Jul 2008, 02:51

1

This post received KUDOS

jimmylow wrote:

If the sequence \(x{_1}\), \(x{_2}\), \(x{_3}\), ...., \(x{_n}\), ....is such that \(x{_1}\) = 3 and \(x{_n_+_1}\) = \(2x{_n} - 1\) for n >= 1, then \(x{_2_0}\) - \(x{_1_9}\) =

A. \(2^{19}\) B. \(2^{20}\) C. \(2^{21}\) D. \(2^{20} -1\) E. \(2^{21} -1\)

A

if you do out the first few, you'll see a pattern.

Re: Math Set 1: Q25 [#permalink]
17 Jul 2008, 03:46

jimmylow wrote:

If the sequence \(x{_1}\), \(x{_2}\), \(x{_3}\), ...., \(x{_n}\), ....is such that \(x{_1}\) = 3 and \(x{_n_+_1}\) = \(2x{_n} - 1\) for n >= 1, then \(x{_2_0}\) - \(x{_1_9}\) =

A. \(2^{19}\) B. \(2^{20}\) C. \(2^{21}\) D. \(2^{20} -1\) E. \(2^{21} -1\)

every term in sequence is 2^n + 1....where n >=1

x20 = 2^20 + 1 x19 = 2^19 + 1

subtracting x19 from x20

2*2^19 + 1 - 2^19 -1

2^19(2-1)

2^19

A.

gmatclubot

Re: Math Set 1: Q25
[#permalink]
17 Jul 2008, 03:46

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