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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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Bunuel
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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] New post   31 Dec 2022, 04:19
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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. \(2^{23}+1\)
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gmatophobia
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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] New post   31 Dec 2022, 06:21
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Bunuel 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. \(2^{23}+1\)


Let's write first few terms of the sequence to establish a pattern

\(A_1=3\)

\(A_2=2.3 - 1\) = \(5\) = \(2^2 + 1\)

\(A_3=2.5 - 1\) = \(9\) = \(2^3 + 1\)

\(A_4=2.9 - 1\) = \(17\) = \(2^4 + 1\)

.
.
.
General Form: \(A_n= 2^n + 1\)

\(A_{23}= 2^{23} + 1\)

\(A_{24}= 2^{24} + 1\)

\(A_{24} - A_{23}= 2^{24} - 2^{23}\)

\(A_{24} - A_{23}= 2^{23} (2-1)\)

\(A_{24} - A_{23}= 2^{23}\)

Option B
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autahh
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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] New post   01 Nov 2023, 03:28
I'm having alot of trouble converting the An+1= 2An-1 to An= 2^n + 1

Is there any background material that can help me go from one to the other?
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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] New post   02 Nov 2023, 07:35
gmatophobia wrote:
Bunuel 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. \(2^{23}+1\)


Let's write first few terms of the sequence to establish a pattern

\(A_1=3\)

\(A_2=2.3 - 1\) = \(5\) = \(2^2 + 1\)

\(A_3=2.5 - 1\) = \(9\) = \(2^3 + 1\)

\(A_4=2.9 - 1\) = \(17\) = \(2^4 + 1\)

.
.
.
General Form: \(A_n= 2^n + 1\)

\(A_{23}= 2^{23} + 1\)

\(A_{24}= 2^{24} + 1\)

\(A_{24} - A_{23}= 2^{24} - 2^{23}\)

\(A_{24} - A_{23}= 2^{23} (2-1)\)

\(A_{24} - A_{23}= 2^{23}\)

Option B



I am a bit confused here...
A1 = 3
so A2 = 2*3 -1
so A3 = 2^2 * 3 - 1
so An = 2^(n-1) *3 - 1

Why are you finding that An = 2^(n) +1 ???
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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]
02 Nov 2023, 07:35
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