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then x20 – x19 =

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then x20 – x19 =

by sanju09 » Fri Oct 08, 2010 4:41 am
If the sequence x1, x2, x3, ..., xn, ... is such that x1 = 3 and x(n + 1) = 2 xn - 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 Rahul@gurome » Fri Oct 08, 2010 5:16 am
x1 = 3
x2 = 2*3 - 1 = 5
x3 = 2*5 - 1 = 9
x4 = 2*9 - 1 = 17 and so on.
Generalizing, we get x(n) = 2^n + 1
So, x(20) = 2^20 + 1
x(19) = 2^19 + 1
x(20) - x(19) = 2^20 + 1 - 2^19 - 1 = 2^20 - 2^19 = 2^19(2 -1) = 2^19

[spoiler]The correct answer is (A).[/spoiler]
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by neilcao » Sat Dec 04, 2010 10:13 pm
Sorry, don't know how to work out the last step: 2^20 - 2^19 = 2^19(2-1)

Why is that? May be a silly questions to ask.
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by ankurmit » Tue Dec 07, 2010 9:23 am
Rahul@gurome wrote:x1 = 3
x2 = 2*3 - 1 = 5
x3 = 2*5 - 1 = 9
x4 = 2*9 - 1 = 17 and so on.
Generalizing, we get x(n) = 2^n + 1
So, x(20) = 2^20 + 1
x(19) = 2^19 + 1
x(20) - x(19) = 2^20 + 1 - 2^19 - 1 = 2^20 - 2^19 = 2^19(2 -1) = 2^19

[spoiler]The correct answer is (A).[/spoiler]
Sorry,I could not get this quistion.

How did you got

x2 = 2*3 - 1 = 5
x3 = 2*5 - 1 = 9
x4 = 2*9 - 1 = 17 and so on?
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by Rahul@gurome » Tue Dec 07, 2010 10:01 am
ankurmit wrote:Sorry,I could not get this quistion.

How did you got

x2 = 2*3 - 1 = 5
x3 = 2*5 - 1 = 9
x4 = 2*9 - 1 = 17 and so on?
x(n + 1) = 2 x(n) - 1 means the (n + 1)th term of the sequence is given by [2*(n-th term of the sequence) - 1].

1st term = x(1) = 3 (Given) = (2^1 + 1)
2nd term = x(2) = (2*x1 - 1) = (2*3 - 1) = 5 = (2^2 + 1)
3rd term = x(3) = (2*x2 - 1) = (2*5 - 1) = 9 = (2^3 + 1)
4th term = x(4) = (2*x3 - 1) = (2*9 - 1) = 17 = (2^4 + 1)
....
n-th term = x(n) = (2^n + 1)

=> 19-th term = x(19) = (2^19 + 1)
=> 20-th term = x(20) = (2^20 + 1)

Hope it is clear now.
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