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There is a sequence An, such that A1=1,A2=2, and An+2=An+1+An+An-1+….+

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There is a sequence An, such that A1=1,A2=2, and An+2=An+1+An+An-1+….+ [#permalink]

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There is a sequence An, such that A1=1,A2=2, and An+2=An+1+An+An-1+….+A2+A1. What is the ratio of A14 to A11?

A. 32
B. 16
C. 8
D. 4
E. 2
[Reveal] Spoiler: OA

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There is a sequence An, such that A1=1,A2=2, and An+2=An+1+An+An-1+….+ [#permalink]

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We know that An+2=An+1+An+An-1+….+A2+A1
Given data: A1=1 and A2=2

From the above information, A3 = A1 + A2 = 1 + 2 = 3
Similarly, A4 = 3 + 2 + 1 = 6

The sequence will be 1,2,3,6,12,24,48,96,192,384,384*2,384*4,384*8,384*16.

The ratio of A14 to A11 is \(\frac{384*16}{384*2}\) = 8(Option C)
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Re: There is a sequence An, such that A1=1,A2=2, and An+2=An+1+An+An-1+….+ [#permalink]

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==> If A11=A10+A9+A8+….+A2+A1=p, you get A12 =A11+A10+A11+….+A2+A1=2p, and from A13=4p, A14=8p, you get A14:A11=8p:p=8:1=8.

The answer is C.
Answer: C
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Re: There is a sequence An, such that A1=1,A2=2, and An+2=An+1+An+An-1+….+ [#permalink]

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New post 08 Sep 2017, 11:17
A(1)=1, A(2)=2, A(3)= 3, A(4)=6, A(5)=12
Seq after n>3 is [2^(n-3)] *3

Let us say x=3
so A(14)= [(2^13)*x]
and A(11)= [(2^8)*x]

ratio would be [(2^13)*x] / [(2^8)*x]
then it would be just [(2^11)]/[(2^8)]= 2^3
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Re: There is a sequence An, such that A1=1,A2=2, and An+2=An+1+An+An-1+….+ [#permalink]

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New post 12 Sep 2017, 19:23
Can you please tell me how derived information for A3

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Re: There is a sequence An, such that A1=1,A2=2, and An+2=An+1+An+An-1+….+ [#permalink]

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New post 12 Sep 2017, 22:52
rghvaggarwal wrote:
Can you please tell me how derived information for A3



We are given that each next term of the seq if formed by adding all the previous terms.

A3= A1+A2
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Re: There is a sequence An, such that A1=1,A2=2, and An+2=An+1+An+An-1+….+   [#permalink] 12 Sep 2017, 22:52
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