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Re: If the sequence x1, x2, x3, …, xn, … is such that x1 = 3 [#permalink]

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19 Sep 2009, 06:07

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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

Soln: the shorter way of doing this is to start from x1, x(1) = 3 Now since x(n+1) = 2x(n) - 1 x(2) = 5 x(3) = 9 x(4) = 17 we can see the pattern now tat x(n) = 2^n + 1 therefore x(20) = 2^20 + 1 x(19) = 2^19 + 1

Hi, Bunnel Your explanation has always been a sigh of high relief to me. Owing to my sluggish expertise, I couldn't figure out as to how you were able to determine that there is a pattern and that is this:

We have the sequence x_1, x_2, x_3, …, x_n,… x_1=3 and x_{n+1}=2x_n - 1 for n\geq1.

Could you please throw light on as to how we should make out the pattern/serious in such questions?

Hi, Bunnel Your explanation has always been a sigh of high relief to me. Owing to my sluggish expertise, I couldn't figure out as to how you were able to determine that there is a pattern and that is this:

We have the sequence x_1, x_2, x_3, …, x_n,… x_1=3 and x_{n+1}=2x_n - 1 for n\geq1.

Could you please throw light on as to how we should make out the pattern/serious in such questions?

Thank you so much.

Well, there is no some kind of general technique to recognize a pattern. The first thing you should do is just write down several terms and look for a pattern.
_________________

I also find it hard to spot patterns specially under time constraints. Sometimes, I write down a couple of items and try to see the difference between each member but it doesn't seem to work that well. If anyone has any suggestions/techniques/tips that could make pattern recognition easier I would be more than grateful

Re: If the sequence x1, x2, x3, …, xn, … is such that x1 = 3 [#permalink]

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24 May 2014, 06:52

srivas 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

Soln: the shorter way of doing this is to start from x1, x(1) = 3 Now since x(n+1) = 2x(n) - 1 x(2) = 5 x(3) = 9 x(4) = 17 we can see the pattern now tat x(n) = 2^n + 1 therefore x(20) = 2^20 + 1 x(19) = 2^19 + 1

so x(20) - x(19) = 2^20 - 2^19 = 2^19(2-1) = 2^19

soln is A

SorryI dont get the last part: 2^20 - 2^19 = 2^(20-19)...which is 2^1?

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

Soln: the shorter way of doing this is to start from x1, x(1) = 3 Now since x(n+1) = 2x(n) - 1 x(2) = 5 x(3) = 9 x(4) = 17 we can see the pattern now tat x(n) = 2^n + 1 therefore x(20) = 2^20 + 1 x(19) = 2^19 + 1

so x(20) - x(19) = 2^20 - 2^19 = 2^19(2-1) = 2^19

soln is A

SorryI dont get the last part: 2^20 - 2^19 = 2^(20-19)...which is 2^1?

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