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There is a sequence an for a positive integer such that a3=a1+a2, a4=

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There is a sequence an for a positive integer such that a3=a1+a2, a4=  [#permalink]

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New post 13 Jan 2017, 01:51
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There is a sequence an for a positive integer such that a3=a1+a2, a4=a3+a2+a1, an =an-1+an-2+an-3+….+a2+a1. If an=p, an+2=?

A. 2p B. 4p C. 8p D. 16p E. 24p

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Re: There is a sequence an for a positive integer such that a3=a1+a2, a4=  [#permalink]

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New post 13 Jan 2017, 02:09
1
MathRevolution wrote:
There is a sequence an for a positive integer such that a3=a1+a2, a4=a3+a2+a1, an =an-1+an-2+an-3+….+a2+a1. If an=p, an+2=?

A. 2p B. 4p C. 8p D. 16p E. 24p


\(a_{n+2} = a_1 + a_2 + a_3 + ... + a_{n-1} + a_n + a_{n+1} = a_n + a_n + (a_n + a_n) = 4a_n = 4p\)

B.
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There is a sequence an for a positive integer such that a3=a1+a2, a4=  [#permalink]

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New post Updated on: 13 Jan 2017, 05:42
\(a_n\) = \(a_{n-1}\) + \(a_{n-2}\) ....+ \(a_1\) ----------------------1

\(a_{n+1}\) = \(a_{n}\) + \(a_{n-1}\) ....+ \(a_1\) =\(a_{n}\) + \(a_{n}\) ----------------------2

\(a_{n+2}\) = \(a_{n+1}\) + \(a_{n}\)+ \(a_{n-1}\) + .... + \(a_1\) ----------------------3

\(a_{n+2}\) = \(a_{n}\) + \(a_{n}\) + \(a_{n}\) + \(a_{n}\) = 4\(a_{n}\) = 4p ( by using st. 1,2 and 3 )

Hence option B.

Originally posted by 0akshay0 on 13 Jan 2017, 02:26.
Last edited by 0akshay0 on 13 Jan 2017, 05:42, edited 1 time in total.
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Re: There is a sequence an for a positive integer such that a3=a1+a2, a4=  [#permalink]

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New post 13 Jan 2017, 04:28
0akshay0 wrote:
\(a_n\) = \(a_{n-1}\) + \(a_{n-2}\) ....+ \(a_1\) ----------------------1

\(a_{n+1}\) = \(a_{n}\) + \(a_{n-1}\) ....+ \(a_1\) =\(a_{n}\) + \(a_{n}\) ----------------------2

\(a_{n+2}\) = \(a_{n+1}\) + \(a_{n}\)+ \(a_{n-1}\) + .... + \(a_1\) ----------------------3

\(a_{n+2}\) = \(a_{n}\) + \(a_{n}\) + \(a_{n}\) + \(a_{n}\) = 4\(a_{n}\) = 4p ( by using st. 1,2 and 3 )



Question is asking what is \(a_{n+2}\)
Given \(a_n\) = \(a_{n-1}\) + \(a_{n-2}\)+\(a_{n-3}\)...........+ \(a_1\) = p

\(a_{n+1}\) = \(a_n\) + \(a_{n-1}\) + \(a_{n-2}\)+\(a_{n-3}\)........... + \(a_1\)
= p + p = 2p

\(a_{n+2}\) = \(a_{n+1}\) + \(a_n\) + \(a_{n-1}\) + \(a_{n-2}\)+\(a_{n-3}\)........... + \(a_1\)
= 2p + p + p = 4p

Answer is B
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There is a sequence an for a positive integer such that a3=a1+a2, a4=  [#permalink]

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New post 13 Jan 2017, 07:34
sb0541 wrote:
0akshay0 wrote:
\(a_n\) = \(a_{n-1}\) + \(a_{n-2}\) ....+ \(a_1\) ----------------------1

\(a_{n+1}\) = \(a_{n}\) + \(a_{n-1}\) ....+ \(a_1\) =\(a_{n}\) + \(a_{n}\) ----------------------2

\(a_{n+2}\) = \(a_{n+1}\) + \(a_{n}\)+ \(a_{n-1}\) + .... + \(a_1\) ----------------------3

\(a_{n+2}\) = \(a_{n}\) + \(a_{n}\) + \(a_{n}\) + \(a_{n}\) = 4\(a_{n}\) = 4p ( by using st. 1,2 and 3 )



Question is asking what is \(a_{n+2}\)
Given \(a_n\) = \(a_{n-1}\) + \(a_{n-2}\)+\(a_{n-3}\)...........+ \(a_1\) = p

\(a_{n+1}\) = \(a_n\) + \(a_{n-1}\) + \(a_{n-2}\)+\(a_{n-3}\)........... + \(a_1\)
= p + p = 2p

\(a_{n+2}\) = \(a_{n+1}\) + \(a_n\) + \(a_{n-1}\) + \(a_{n-2}\)+\(a_{n-3}\)........... + \(a_1\)
= 2p + p + p = 4p

Answer is B



I did the same thing.
St. 1,2 and 3 explains how \(a_{n+2}\) = 4\(a_n\) :)
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Re: There is a sequence an for a positive integer such that a3=a1+a2, a4=  [#permalink]

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New post 15 Jan 2017, 18:45
==> Since an =an-1+an-2+an-3+….+a2+a1 =p,
an+1 =an+ an-1+an-2+an-3+….+a2+a1=p+p=2p
an+2=an+1+an+an-1+an-2+an-3+….+a2+a1=2p+p+p=4p.

Hence, the answer is B.
Answer: B
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Re: There is a sequence an for a positive integer such that a3=a1+a2, a4=  [#permalink]

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