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If the sequence x1, x2,x3,...,xn is such that X1=3 and Xn+1 [#permalink ]
06 May 2008, 11:05

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

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hmm it took me a long time ... but lets try this ...

start writing out some of the resultants and should start seeing a pattern where each value corresponds to 2^xn + 1 ... therefore :

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

therefore the answer is A

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puma 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

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My answer is A 2^19 Xn+1 = 2Xn -1 ,X1=3 X2 = 3*2 -1 = 5 (X2-X1 = 2) 2^1 X3 =5*2 -1 = 9 (X3-X2 =4 ) 2^2 X4 = 9*2 -1 =17 (X4-X3 = 8 ) 2^3 So we have a pattern where Xn - Xn-1 =2^n-1 Therefore for X20 -X19 = 2^19

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puma 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

your post is not very clear..however..here is how i do it..

xn+1=(2xn)-1

x1=3, x2=5, x3=9 x4=17 x5=33

notice the difference btw x^5-x^4=2^4 or 16, x^4-x^3=2^3...

therefore x^20-x^19=2^19

A it is..

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puma 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

X1 = 3 = 2^1 + 1

X2 = 5 = 2^2 + 1

X3 = 9 = 2^3 + 1

This means

x19 = 2^19 + 1

X20 = 2^20 + 1

X20 - X19 = 2^20 + 1 - 2^19 - 1 = 2^20 - 2^19 = 2^19*(2-1) = 2^19

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I got A X2=2X1 -1=5=2^2+1 x3=2x2-1=10-1=2^3+1 ... x20-x19= 2x19-1-x19=x19-1 but x19=2^19+1 thus x19-1= 2^19+1-1=2^19

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