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PS: series

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Director
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Joined: 01 Apr 2008
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Name: Ronak Amin
Schools: IIM Lucknow (IPMX) - Class of 2014
PS: series [#permalink]

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New post 24 Mar 2009, 01:15
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A
B
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D
E

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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

Note: xn is the nth term, 2xn is 2*nth term. I dont know how to write this notation here.

Kudos [?]: 860 [0], given: 18

SVP
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Joined: 29 Aug 2007
Posts: 2471

Kudos [?]: 856 [0], given: 19

Re: PS: series [#permalink]

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New post 24 Mar 2009, 05:27
Economist wrote:
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

Note: xn is the nth term, 2xn is 2*nth term. I dont know how to write this notation here.

x1 = 3
x(n+1) = 2x(n) - 1
x(2) = 2x(1) - 1 = 2x3 - 1 = 2^2 + 1
x(3) = 2x(2) - 1 = 2x5 - 1 = 2^3 + 1
x(4) = 2x(3) - 1 = 2x9 - 1 = 2^4 + 1
x(5) = 2x(4) - 1 = 2x17 - 1 = 2^5 +1

x(19) = 2^19 +1
x(20) = 2^20 +1

x20 – x19 = (2^20 +1) - (2^19 +1) = 2^20 +1 - 2^19 - 1 = 2^19 (2-1) = 2^20

A.
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Kudos [?]: 856 [0], given: 19

Intern
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Joined: 07 Feb 2009
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Re: PS: series [#permalink]

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New post 24 Mar 2009, 06:23
hi GMAT TIGER,

Great solution.
But i feel that there is some calculation mistake in the last line, which gives a wrong result.

The answer should be A instead of B.

Here it goes:

As u have derived it,
x(19) = 2^19 +1
x(20) = 2^20 +1

x20 – x19 = (2^20 +1) - (2^19 +1) = 2^19 (2-1) = 2^19.

Tell me the OA plz.

Kudos [?]: 17 [0], given: 1

Director
Director
User avatar
Joined: 01 Apr 2008
Posts: 872

Kudos [?]: 860 [0], given: 18

Name: Ronak Amin
Schools: IIM Lucknow (IPMX) - Class of 2014
Re: PS: series [#permalink]

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New post 24 Mar 2009, 06:25
Hi
Didnt understand the last part:

x20 – x19 = (2^20 +1) - (2^19 +1) = 2^19 (3-1) = 2^20
it shud be
2^20 - 2^19 = 2^19 (2-1) = 2^19.

GMAT TIGER wrote:
Economist wrote:
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

Note: xn is the nth term, 2xn is 2*nth term. I dont know how to write this notation here.

x1 = 3
x(n+1) = 2x(n) - 1
x(2) = 2x(1) - 1 = 2x3 - 1 = 2^2 + 1
x(3) = 2x(2) - 1 = 2x5 - 1 = 2^3 + 1
x(4) = 2x(3) - 1 = 2x9 - 1 = 2^4 + 1
x(5) = 2x(4) - 1 = 2x17 - 1 = 2^5 +1

x(19) = 2^19 +1
x(20) = 2^20 +1

x20 – x19 = (2^20 +1) - (2^19 +1) = 2^19 (3-1) = 2^20

B.

Kudos [?]: 860 [0], given: 18

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Re: PS: series [#permalink]

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New post 24 Mar 2009, 09:01
x20=2 *x19 -11
hence x20-x19 = x19 - 1 ------------------>(1)
Now x2= 5 = 2^2+1
x3=9= 2^3 +1
Generalising, x(n)= 2^n + 1
Hence x(19)= 2^19 +1
Put in (1),

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

Hence A

Kudos [?]: 541 [0], given: 0

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Kudos [?]: 856 [0], given: 19

Re: PS: series [#permalink]

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New post 24 Mar 2009, 16:43
abhishekik wrote:
hi GMAT TIGER,

Great solution.
But i feel that there is some calculation mistake in the last line, which gives a wrong result.

The answer should be A instead of B.

Here it goes:

As u have derived it,
x(19) = 2^19 +1
x(20) = 2^20 +1

x20 – x19 = (2^20 +1) - (2^19 +1) = 2^19 (2-1) = 2^19.

Tell me the OA plz.


Thanks. I updated.
_________________

Verbal: http://gmatclub.com/forum/new-to-the-verbal-forum-please-read-this-first-77546.html
Math: http://gmatclub.com/forum/new-to-the-math-forum-please-read-this-first-77764.html
Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html


GT

Kudos [?]: 856 [0], given: 19

Re: PS: series   [#permalink] 24 Mar 2009, 16:43
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