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Manager
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The sequence a1, a2, a3, ...an is such that a2 = a1, a3 = [#permalink]
 26 Jan 2008, 00:08
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The sequence a1, a2, a3, ...an is such that a2 = a1, a3 = a1+a2,
an = a1 + a2 + a3......+a(n-1). If an = k, where n >2, a(n+3) =?
A) 2k
B) 4k
C) 6k
D) 8k
E) 16k
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CEO
Joined: 17 Nov 2007
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Concentration: Entrepreneurship, Other
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Da_n = a_1 + a_2 + a_3......+a_{n-1}a_{n+1} = a_1 + a_2 + a_3......+a_{n-1}+a_n=a_n+a_n=2a_na_{n+m} =2^m*a_na_{n+3} =2^3*k=8k
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Senior Manager
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walker wrote: D
a_n = a_1 + a_2 + a_3......+a_{n-1}
a_{n+1} = a_1 + a_2 + a_3......+a_{n-1}+a_n=a_n+a_n=2a_n
a_{n+m} =2^m*a_n
a_{n+3} =2^3*k=8k D too. We know that: a2= a1 a3= a2+(a1)=a2+a2= 2a2 a4=a3+(a2+a1)=a3+a3=2a3 That is, we have that an=2an-1 Besides, we know that an=k Then, an+3=2an+2=4an+1=8an=8k
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Manager
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Yes. 8k is the answer. :D
Thank you for your explanation.
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Director
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a1, a2, (a1+a2), 2*(a1+a2), 4*(a1+a2), 8*(a1+a2),...
For example: a3 = (a1+a2) = k
a(3+3)= a6 = 8*(a1+a2) = 8k
"D"
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CEO
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GHIBI wrote: The sequence a1, a2, a3, ...an is such that a2 = a1, a3 = a1+a2,
an = a1 + a2 + a3......+a(n-1). If an = k, where n >2, a(n+3) =?
A) 2k
B) 4k
C) 6k
D) 8k
E) 16k D. an = k a(n+1) = 2k a(n+2) = 2 x 2k = 4k a(n+3) = 2 x 4k = 8k
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