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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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New post 05 Jun 2016, 06:04
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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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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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New post 05 Jun 2016, 08:47
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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
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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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New post 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

\(A(24) - A(23) = 2^{22}[A2 - A1]\)
\(\therefore A(24) - A(23) = 2^{22}[5 - 3 ] = 2^{23}\)
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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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New post 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.

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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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New post 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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New post 12 Apr 2017, 18:19
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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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New post 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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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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