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If the sequence x1, x2, x3, …, xn, … is such that x1 = 3

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

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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: if-the-sequence-x1-x2-x3-xn-is-such-that-x1-98536.html
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Re: If the sequence x1, x2, x3, …, xn, … is such that x1 = 3 [#permalink]

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New post 17 Sep 2009, 00:29
x20=2.x19-1
x20-x19=2.x19-1-x19=x19-1................1
x19=2.x18-1..............2
putting in 2 in 1
x20-x19=2.x18-2...........3
x18=2.x17-1-1..............4
putting 4 in 3
x20-x19=2.(2.x17-1)-1-1=4.x17-2-1-1
....
....
x2=2.x1-1
x20-x19=2^18.x1-(2^17+2^16+......+2+1+1)
x20-x19=2^18.3-2^18=2^19
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Re: If the sequence x1, x2, x3, …, xn, … is such that x1 = 3 [#permalink]

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New post 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 [#permalink]

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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

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

soln is A
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PS q25 from set 1 (31's sets) [#permalink]

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New post 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) [#permalink]

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New post 29 Oct 2009, 16:07
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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 \(x_19=2^{19}+1\).

\(x_{20}-x_{19}=2^{20}+1-2^{19}-1=2^{20}-2^{19}=2^{19}\)

Answer: A.

Hope it helps.
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Re: PS q25 from set 1 (31's sets) [#permalink]

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New post 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) [#permalink]

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New post 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 \(x_19=2^{19}+1\).

\(x_{20}-x_{19}=2^{20}+1-2^{19}-1=2^{20}-2^{19}=2^{19}\)

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) [#permalink]

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New post 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 \(x_19=2^{19}+1\).

\(x_{20}-x_{19}=2^{20}+1-2^{19}-1=2^{20}-2^{19}=2^{19}\)

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 [#permalink]

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New post 19 May 2012, 19:27
Benelux, you may need to show how you factored out 2^19 and multiplied (2-1). Knowing it and seeing it was the difference for me.
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Re: PS q25 from set 1 (31's sets) [#permalink]

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New post 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 [#permalink]

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New post 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 \(r^{n-1}\)

2 x \(2^{n-1}\)

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 [#permalink]

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

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New post 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(2-1) = 2^19

soln is A



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


\(2^{20}-2^{19}\) --> factor out \(2^{19}\):

\(2^{20}-2^{19}=2^{19}(2-1)= 2^{19}\).

Hope it's clear.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-the-sequence-x1-x2-x3-xn-is-such-that-x1-98536.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   [#permalink] 24 May 2014, 05:57
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