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# There is a sequence A(n) such that A(n+1)=2A(n)-1 and A(1)=3, where n

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There is a sequence A(n) such that A(n+1)=2A(n)-1 and A(1)=3, where n  [#permalink]

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05 Jun 2016, 06:04
2
9
00:00

Difficulty:

45% (medium)

Question Stats:

64% (02:35) correct 36% (03:09) wrong based on 119 sessions

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There is a sequence A(n) such that A(n+1)=2A(n)-1 and A(1)=3, where n is a positive integer. What is the value of A(24)-A(23)?
A. 2^22+1
B. 2^23
C. 2^23+1
D. 2^24
E. 2^23+1

*An answer will be posted in 2 days.

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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" ##### Most Helpful Community Reply Current Student Joined: 18 Oct 2014 Posts: 803 Location: United States GMAT 1: 660 Q49 V31 GPA: 3.98 Re: There is a sequence A(n) such that A(n+1)=2A(n)-1 and A(1)=3, where n [#permalink] ### Show Tags 05 Jun 2016, 08:47 3 2 MathRevolution wrote: There is a sequence A(n) such that A(n+1)=2A(n)-1 and A(1)=3, where n is a positive integer. What is the value of A(24)-A(23)? A. 2^22+1 B. 2^23 C. 2^23+1 D. 2^24 E. 2^23+1 *An answer will be posted in 2 days. A1= 3 A2= 2*3-1= 5 A3= 2*5-1= 9 A4= 2*9-1= 17 We can notice that there is a squence A2-A1= 2^1 A3-A2= 2^2 A4-A3= 2^3 Hence A24-A23= 2^23 B is the answer _________________ I welcome critical analysis of my post!! That will help me reach 700+ ##### General Discussion Manager Joined: 28 Aug 2006 Posts: 248 Re: There is a sequence A(n) such that A(n+1)=2A(n)-1 and A(1)=3, where n [#permalink] ### Show Tags 05 Jun 2016, 07:54 Couple of ways to answer this .... Method 1: Find the pattern Let's say A(1) = x for a while Now, let's write the series bearing in mind that each term is 1 less than twice its preceding term. $$A1 = x$$ $$A2 = 2x -1$$ $$A3 = 4x -3$$ $$A4 = 8x -7$$ $$A5 = 16x -15$$ $$A6 = 32x -31$$ ............. ............. Notice the following $$A2 - A1 = x - 1$$ $$A3 - A2 = 2(x-1) = 2^1(x-1)$$ $$A4 - A3 = 4(x-1) = 2^2(x-1)$$ $$A5 - A4 = 8(x-1) = 2^3(x-1)$$ $$A6 - A5 = 16(x-1) = 2^4(x-1)$$ $$=> A24 - A23 = 2^{22}(x-1)$$ $$=> A24 - A23 = 2^{22}(3-1) = 2^{23}$$ (Given x = 3 in the question) $$=> A24 - A23 = 2^{23}$$ Method 2: Going to the start of the series Given $$A(n+1) = 2A(n) - 1$$ $$=> A(n+1) + 1 = 2A(n)$$ --------(i) Bearing this format in mind, let us write the given question the following way $$A(24) - A(23)$$ $$= A(24) + 1 - A(23) - 1$$ $$= [A(24) +1 ] - [A(23) + 1]$$ $$= 2A(23) - 2A(22)$$ ( From (i) ) $$= 2[A(23) - A(22)]$$ Similarly $$A(23) - A(22)$$ will equal $$2[A(22) - A(21)]$$ $$\therefore A(24) - A(23) = 2^2[A(22) - A(21)]$$ Continuing this, we get $$A(24) - A(23) = 2^{22}[A2 - A1]$$ $$\therefore A(24) - A(23) = 2^{22}[5 - 3 ] = 2^{23}$$ _________________ Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 8017 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: There is a sequence A(n) such that A(n+1)=2A(n)-1 and A(1)=3, where n [#permalink] ### Show Tags 06 Jun 2016, 21:55 From A(1)=3=2+1, A(2)=2A(1)-1=5=2^2+1,…A(n)=2^n+1, we can calculate A(24)-A(23)=(2^24+1)-(2^23-1)=2^24-2^23=(2-1)2^23=2^23. Hence, the correct answer is B. _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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Re: There is a sequence A(n) such that A(n+1)=2A(n)-1 and A(1)=3, where n  [#permalink]

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29 Mar 2017, 00:58
I missed it because i started iteration by A(24) - A(23) and tried to decrease it.

It is better to start with A(2) - A(1), increase interations and infer A(24)-A(23).

I like the question!!!
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Re: There is a sequence A(n) such that A(n+1)=2A(n)-1 and A(1)=3, where n  [#permalink]

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12 Apr 2017, 18:19
1
MathRevolution wrote:
There is a sequence A(n) such that A(n+1)=2A(n)-1 and A(1)=3, where n is a positive integer. What is the value of A(24)-A(23)?
A. 2^22+1
B. 2^23
C. 2^23+1
D. 2^24
E. 2^23+1

*An answer will be posted in 2 days.

Working out the first few terms gives:

A1=3
A2=5
A3=9
A4=17
A5=33

The differences between each of the terms are 2, 4, 8, 16. Each term increases by 2 to an increased power. A1 increases by 2^1, A2 increases by 2^2 and A3 increases by 2^3 and so on.
So A23 will increase by 2^23 and that's the difference.

Is this the correct method? Is there a better one?
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Re: There is a sequence A(n) such that A(n+1)=2A(n)-1 and A(1)=3, where n  [#permalink]

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18 Apr 2017, 07:26
Option B

A(n+1) = 2A(n) - 1; A(1) = 3.

A(2) = 2*3 - 1 = 5 = 2^2 + 1
A(3) = 2*5 - 1 = 9 = 2^3 + 1
A(4) = 2*9 - 1 = 17 = 2^4 + 1
A(5) = 2*17 - 1 = 33 = 2^5 + 1
.
.
A(23) = 2^23 + 1
A(24) = 2^24 + 1

A(24) - A(23) = 2^23(2-1) = 2^23
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Re: There is a sequence A(n) such that A(n+1)=2A(n)-1 and A(1)=3, where n  [#permalink]

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19 Sep 2018, 22:52
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Re: There is a sequence A(n) such that A(n+1)=2A(n)-1 and A(1)=3, where n   [#permalink] 19 Sep 2018, 22:52
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