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If the sequence x1, x2, x3, …, xn, … is such that x1 = 3
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16 Sep 2009, 23:05
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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 OPEN DISCUSSION OF THIS QUESTION IS HERE: ifthesequencex1x2x3xnissuchthatx198536.html
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Re: If the sequence x1, x2, x3, …, xn, … is such that x1 = 3
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17 Sep 2009, 00:29
x20=2.x191 x20x19=2.x191x19=x191................1 x19=2.x181..............2 putting in 2 in 1 x20x19=2.x182...........3 x18=2.x1711..............4 putting 4 in 3 x20x19=2.(2.x171)11=4.x17211 .... .... x2=2.x11 x20x19=2^18.x1(2^17+2^16+......+2+1+1) x20x19=2^18.32^18=2^19



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Re: If the sequence x1, x2, x3, …, xn, … is such that x1 = 3
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17 Sep 2009, 05:00
I wud go with option A) 2^ 19



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Re: If the sequence x1, x2, x3, …, xn, … is such that x1 = 3
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19 Sep 2009, 05:07
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(21) = 2^19
soln is A



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PS q25 from set 1 (31's sets)
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29 Oct 2009, 10:53
Hi! I've just started my practice phase of GMAT preparation and was very surprised to get wrong answer on the question that I was sured I did right. If the sequence \(x_1, x_2, x_3,\) …, \(x_n,\)… is such that \(x_1=3\) and \(x_{n+1}=2x_n  1\) for \(n\geq1\), then \(x_{20}  x_{19} =\) A. \(2^{19}\) B. \(2^{20}\) C. \(2^{21}\) D. \(2^{20}1\) E. \(2^{21}1\)
The OA is A, but I answered B and don't understand what I've missed. P.s. Pls, advise me on how to make powers a little bit smaller?
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Re: PS q25 from set 1 (31's sets)
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29 Oct 2009, 16:07
Vyacheslav wrote: Hi! I've just started my practice phase of GMAT preparation and was very surprised to get wrong answer on the question that I was sured I did right. If the sequence \(x_1, x_2, x_3,\) …, \(x_n,\)… is such that \(x_1=3\) and \(x_{n+1}=2x_n  1\) for \(n\geq1\), then \(x_{20}  x_{19} =\) A. \(2^{19}\) B. \(2^{20}\) C. \(2^{21}\) D. \(2^{20}1\) E. \(2^{21}1\)
The OA is A, but I answered B and don't understand what I've missed. P.s. Pls, advise me on how to make powers a little bit smaller? We have the sequence \(x_1, x_2, x_3,\) …, \(x_n,\)… \(x_1=3\) and \(x_{n+1}=2x_n  1\) for \(n\geq1\). If you notice there is a specific pattern int it: \(x_1=3=2^1+1\) \(x_2=5=2^2+1\) \(x_3=9=2^3+1\) ... \(x_n=2^n+1\) So, \(x_{20}=2^{20}+1\) and \(x_19=2^{19}+1\). \(x_{20}x_{19}=2^{20}+12^{19}1=2^{20}2^{19}=2^{19}\) Answer: A. Hope it helps.
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Re: PS q25 from set 1 (31's sets)
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31 Oct 2009, 04:19
I was trying to do it by another way: x20  x19 = x19  1. Now, x19 = 2 (x18)  1 = 2 ( 2 (x17  1 ) )  1 and so on... we get 2^18(x1  1 )  1 x19 = 2^19  1 Hence, required ans is 2^19  1  1...I am missing something for sure



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Re: PS q25 from set 1 (31's sets)
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17 May 2012, 22:00
Bunuel wrote: Vyacheslav wrote: Hi! I've just started my practice phase of GMAT preparation and was very surprised to get wrong answer on the question that I was sured I did right. If the sequence \(x_1, x_2, x_3,\) …, \(x_n,\)… is such that \(x_1=3\) and \(x_{n+1}=2x_n  1\) for \(n\geq1\), then \(x_{20}  x_{19} =\) A. \(2^{19}\) B. \(2^{20}\) C. \(2^{21}\) D. \(2^{20}1\) E. \(2^{21}1\)
The OA is A, but I answered B and don't understand what I've missed. P.s. Pls, advise me on how to make powers a little bit smaller? We have the sequence \(x_1, x_2, x_3,\) …, \(x_n,\)… \(x_1=3\) and \(x_{n+1}=2x_n  1\) for \(n\geq1\). If you notice there is a specific pattern int it: \(x_1=3=2^1+1\) \(x_2=5=2^2+1\) \(x_3=9=2^3+1\) ... \(x_n=2^n+1\) So, \(x_{20}=2^{20}+1\) and \(x_19=2^{19}+1\). \(x_{20}x_{19}=2^{20}+12^{19}1=2^{20}2^{19}=2^{19}\) Answer: A. Hope it helps. 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.



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Re: PS q25 from set 1 (31's sets)
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18 May 2012, 00:13
smileforever41 wrote: Bunuel wrote: Vyacheslav wrote: Hi! I've just started my practice phase of GMAT preparation and was very surprised to get wrong answer on the question that I was sured I did right. If the sequence \(x_1, x_2, x_3,\) …, \(x_n,\)… is such that \(x_1=3\) and \(x_{n+1}=2x_n  1\) for \(n\geq1\), then \(x_{20}  x_{19} =\) A. \(2^{19}\) B. \(2^{20}\) C. \(2^{21}\) D. \(2^{20}1\) E. \(2^{21}1\)
The OA is A, but I answered B and don't understand what I've missed. P.s. Pls, advise me on how to make powers a little bit smaller? We have the sequence \(x_1, x_2, x_3,\) …, \(x_n,\)… \(x_1=3\) and \(x_{n+1}=2x_n  1\) for \(n\geq1\). If you notice there is a specific pattern int it: \(x_1=3=2^1+1\) \(x_2=5=2^2+1\) \(x_3=9=2^3+1\) ... \(x_n=2^n+1\) So, \(x_{20}=2^{20}+1\) and \(x_19=2^{19}+1\). \(x_{20}x_{19}=2^{20}+12^{19}1=2^{20}2^{19}=2^{19}\) Answer: A. Hope it helps. 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.
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Re: If the sequence x1, x2, x3, …, xn, … is such that x1 = 3
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19 May 2012, 19:27
Benelux, you may need to show how you factored out 2^19 and multiplied (21). Knowing it and seeing it was the difference for me.



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Re: PS q25 from set 1 (31's sets)
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05 Oct 2013, 12:47
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
Cheers!



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Re: If the sequence x1, x2, x3, …, xn, … is such that x1 = 3
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07 Feb 2014, 20:33
No frills or shortcuts, just straight algebra.
For \(x_n\) = 2\(x_n\)  1:
\(x_1\)= 3
+2
\(x_2\)= 5
+4
\(x_3\)= 9
+8
\(x_4\)= 17
+16
\(x_5\)= 33,
Notice the difference between each entry of the sequence is a geometric progression: 2, 4, 8, 16.
\(x_1\) x \(r^{n1}\)
2 x \(2^{n1}\)
Thus,
= \(x_{20}\)  \(x_{19}\)
= (2 x \(2^{19}\))  (2 x \(2^{18}\))
= 2 x \(2^{18}\) (2  1)
= \(2^{19}\) (1)
= \(2^{19}\)



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Re: If the sequence x1, x2, x3, …, xn, … is such that x1 = 3
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24 May 2014, 05:52
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(21) = 2^19
soln is A SorryI dont get the last part: 2^20  2^19 = 2^(2019)...which is 2^1?



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Re: If the sequence x1, x2, x3, …, xn, … is such that x1 = 3
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24 May 2014, 05:57
usre123 wrote: 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(21) = 2^19
soln is A SorryI dont get the last part: 2^20  2^19 = 2^(2019)...which is 2^1? \(2^{20}2^{19}\) > factor out \(2^{19}\): \(2^{20}2^{19}=2^{19}(21)= 2^{19}\). Hope it's clear. OPEN DISCUSSION OF THIS QUESTION IS HERE: ifthesequencex1x2x3xnissuchthatx198536.html
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: If the sequence x1, x2, x3, …, xn, … is such that x1 = 3
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Re: If the sequence x1, x2, x3, …, xn, … is such that x1 = 3
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