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If the sequence X1, X2, X3, , Xn, is such thatX1 = 3 andXn+1 [#permalink]
 12 Nov 2008, 08:39
If the sequence X1, X2, X3, …, Xn, … is such thatX1 = 3 andXn+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
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Re: sequence number --27 [#permalink]
12 Nov 2008, 10:08
please repost your question..i dont know if it 2Xn - 1 or 2X(n-1)
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Current Student
Joined: 28 Dec 2004
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Re: sequence number --27 [#permalink]
12 Nov 2008, 10:31
anyways assuming it reads X(n+1)=2(Xn)-1 then
x(2)=2(x1)-1 or 5 or (2^n) +1
x(3)=2(5)-1 or 9 (2^n)+1
x(4)=2(9)-1 or 17 (2^n)+1
x(3)-x(4)=[(2^3)+1]-[2^4+1]= 2^3-2^4 or 2^3(2-1) or 2^3
x(20)-x(19)=2^19
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Re: sequence number --27 [#permalink]
12 Nov 2008, 10:50
I have not spotted (2^n) +1 pattern until I saw fresinha12's response.
However, once we spot this pattern the solution could be as follows
x(20)-x(19)
2(x(19))-1-x(19)
x(19)-1
2^19+1-1=2^19
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Re: sequence number --27
[#permalink]
12 Nov 2008, 10:50
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