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sequence

by roger federer » Thu Mar 11, 2010 2:32 pm
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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by sanju09 » Fri Mar 12, 2010 12:34 am
roger federer 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
When the rule is followed, the sequence appears to be running like...

3, 5, 9, 17, 33, ...

noticed anything?

Ist and IInd differ by 2^1, IInd and IIIrd differ by 2^2, IIIrd and IVth differ by 2^3,...and so on, to let us conclude that 19th and 20th would differ by [spoiler]2^19.

A[/spoiler]
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by kstv » Fri Mar 12, 2010 5:46 am
If the sequence x1, x2, x3, ..., xn, ... is such that x1 = 3 and xn+1 = 2xn - 1 for n ≥ 1, then x20 -
x19 =
x20 = x19+1 = 2x19-1
x20-x19 = 2x19+1-x19=x19
x1 = 3 x2=2x1-1 = 6-1= 5 x3=2x2-1=9 x4=2x3-1=17 or x4= 2[2{2( x1)-1)}-1)]-1=2(2(2*3)-1]-1)-1)
= 2^18

Too tedious to carry on , not the method to solve in 2 min

A. 2^19
B. 2^20
C. 2^21
D. 2^20 - 1
E. 2^21 - 1
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