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A certain computer program generates a sequence of numbers

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A certain computer program generates a sequence of numbers [#permalink]

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New post Updated on: 11 Dec 2013, 01:45
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A certain computer program generates a sequence of numbers a1, a2, … , an such that a1 = a2 = 1 and ak = ak-1 + 2ak-2 for all integers k such that 3 ≤ k ≤ n. If n > 6, then a7 = ?

A. 32
B. 43
C. 64
D. 100
E. 128

Originally posted by Nihit on 13 Sep 2008, 07:56.
Last edited by Bunuel on 11 Dec 2013, 01:45, edited 1 time in total.
Added the OA.
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Re: sequence [#permalink]

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New post 13 Sep 2008, 09:36
2
Nihit wrote:
A certain computer program generates a sequence of numbers a1, a2, … , an such that a1 = a2 = 1 and ak = ak-1 + 2ak-2 for all integers k such that 3 ≤ k ≤ n. If n > 6, then a7 = ?

A. 32
B. 43
C. 64
D. 100
E. 128



a1=1
a2=1
a3=a2+2a1= 3
a4=a3+2a2=5
a5=11
a6=21
a7=43

B
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Re: sequence [#permalink]

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New post 14 Sep 2008, 23:05
another B

just for convenience I will explain lil bit more

Ak= A(k-1) + 2* A(k-2)

in other words every A in this sequence is the sum of pervious term and twice the term that is prev. to prev.....lol it would be easier with working

A1= 1 and A2= 1

A(3)= A(3-1) + A(3-2)
=> = A(2) + 2(A1)
=> = 1 + 2*1
A(3)= 3
And with similar pattern,

A(4)= 3 + 2*1 = 5
A(5)= 5 + 2*3 = 11
A(6)= 11 + 2*5 = 21
A(7)= 21 + 2*11 = 43
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Re: sequence [#permalink]

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New post 01 Mar 2010, 07:10
x2suresh wrote:
Nihit wrote:
A certain computer program generates a sequence of numbers a1, a2, … , an such that a1 = a2 = 1 and ak = ak-1 + 2ak-2 for all integers k such that 3 ≤ k ≤ n. If n > 6, then a7 = ?

A. 32
B. 43
C. 64
D. 100
E. 128



a1=1
a2=1
a3=a2+2a1= 3
a4=a3+2a2=5
a5=11
a6=21
a7=43

B


Liked your approach.
I solved by breaking up a7 until I reached a1 and a2. But it took me while to reach the correct answer.
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Re: sequence [#permalink]

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New post 03 Mar 2010, 10:12
Nihit wrote:
A certain computer program generates a sequence of numbers a1, a2, … , an such that a1 = a2 = 1 and ak = ak-1 + 2ak-2 for all integers k such that 3 ≤ k ≤ n. If n > 6, then a7 = ?

A. 32
B. 43
C. 64
D. 100
E. 128


it's B.
a3= a2+a1 = 3
a4 = 3+2 = 5
a5 = 5+6=11
a6=11+10=21
a7=21+22=43

This may be a better way to solve this as there no constant difference between two consecutive terms like in AP.
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Re: sequence [#permalink]

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New post 03 Mar 2010, 11:53
11
Nihit wrote:
A certain computer program generates a sequence of numbers a1, a2, … , an such that a1 = a2 = 1 and ak = ak-1 + 2ak-2 for all integers k such that 3 ≤ k ≤ n. If n > 6, then a7 = ?

A. 32
B. 43
C. 64
D. 100
E. 128

Solving as mentioend earlier always helps but here's another approach for this particular problem
a1 and a2 are odd and a3 would be odd and all the numbers in the sequence are odd as they are of the format
odd number + 2 * odd number.
a7 is odd and only B has an odd number

B
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Re: sequence [#permalink]

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New post 28 Jun 2011, 11:51
43...choice B

a5=11; a6=21; a7=43
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Re: sequence [#permalink]

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New post 29 Jun 2011, 17:06
a1 = a2 =1
a3 = a2+2a1 = 3
a4 = a3+2a2 = 5
a5 = a4+2a3 = 11
a6 = a5+2a4 = 21
a7 = a6+2a5 = 43

Answer is B.
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Re: sequence [#permalink]

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New post 29 Jun 2011, 17:08
even though this is good observation. this approach wont work if there are more than 1 odd number in the answer choices.

chix475ntu wrote:
Nihit wrote:
A certain computer program generates a sequence of numbers a1, a2, … , an such that a1 = a2 = 1 and ak = ak-1 + 2ak-2 for all integers k such that 3 ≤ k ≤ n. If n > 6, then a7 = ?

A. 32
B. 43
C. 64
D. 100
E. 128

Solving as mentioend earlier always helps but here's another approach for this particular problem
a1 and a2 are odd and a3 would be odd and all the numbers in the sequence are odd as they are of the format
odd number + 2 * odd number.
a7 is odd and only B has an odd number

B
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Re: sequence [#permalink]

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New post 17 Feb 2014, 13:24
x2suresh wrote:
Nihit wrote:
A certain computer program generates a sequence of numbers a1, a2, … , an such that a1 = a2 = 1 and ak = ak-1 + 2ak-2 for all integers k such that 3 ≤ k ≤ n. If n > 6, then a7 = ?

A. 32
B. 43
C. 64
D. 100
E. 128



a1=1
a2=1
a3=a2+2a1= 3
a4=a3+2a2=5
a5=11
a6=21
a7=43

B


Why do they give you this stuff? '3 ≤ k ≤ n. If n > 6'?

Thanks
Cheers
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Re: sequence [#permalink]

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New post 17 Feb 2014, 21:29
jlgdr wrote:
x2suresh wrote:
Nihit wrote:
A certain computer program generates a sequence of numbers a1, a2, … , an such that a1 = a2 = 1 and ak = ak-1 + 2ak-2 for all integers k such that 3 ≤ k ≤ n. If n > 6, then a7 = ?

A. 32
B. 43
C. 64
D. 100
E. 128



a1=1
a2=1
a3=a2+2a1= 3
a4=a3+2a2=5
a5=11
a6=21
a7=43

B


Why do they give you this stuff? '3 ≤ k ≤ n. If n > 6'?

Thanks
Cheers
J


Because the relation they gave you \(a_k = a_{k-1} + 2a_{k-2}\) holds only from the third term onwards till the last term \(a_n\). That is, k should be 3 or more and will take the last value of n. This relation doesn't hold for the first and second terms and holds for all the rest of the terms. That is what this 'stuff' specifies.

If n > 6 tells you that the sequence has more than 6 terms i.e. at least 7 terms.
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Re: A certain computer program generates a sequence of numbers [#permalink]

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New post 03 Jul 2014, 09:58
rishi2377 wrote:
another B

just for convenience I will explain lil bit more

Ak= A(k-1) + 2* A(k-2)

in other words every A in this sequence is the sum of pervious term and twice the term that is prev. to prev.....lol it would be easier with working

A1= 1 and A2= 1

A(3)= A(3-1) + A(3-2)
=> = A(2) + 2(A1)
=> = 1 + 2*1
A(3)= 3
And with similar pattern,

A(4)= 3 + 2*1 = 5
A(5)= 5 + 2*3 = 11
A(6)= 11 + 2*5 = 21
A(7)= 21 + 2*11 = 43


why isn't the 2 being multiplied when you do this : A(3)= A(3-1) + A(3-2) isn't the rule given :A(k-1) + 2* A(k-2)
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Re: A certain computer program generates a sequence of numbers [#permalink]

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New post 03 Jul 2014, 19:13
sagnik2422 wrote:
rishi2377 wrote:
another B

just for convenience I will explain lil bit more

Ak= A(k-1) + 2* A(k-2)

in other words every A in this sequence is the sum of pervious term and twice the term that is prev. to prev.....lol it would be easier with working

A1= 1 and A2= 1

A(3)= A(3-1) + A(3-2)
=> = A(2) + 2(A1)
=> = 1 + 2*1
A(3)= 3
And with similar pattern,

A(4)= 3 + 2*1 = 5
A(5)= 5 + 2*3 = 11
A(6)= 11 + 2*5 = 21
A(7)= 21 + 2*11 = 43


why isn't the 2 being multiplied when you do this : A(3)= A(3-1) + A(3-2) isn't the rule given :A(k-1) + 2* A(k-2)


That's just a typo. Note the next step of the poster: A(3) = A(2) + 2(A1)
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Re: A certain computer program generates a sequence of numbers [#permalink]

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New post 19 Jan 2015, 01:20
calculating in sequence

1,1,3,5,11,21,43

Ans B
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Re: A certain computer program generates a sequence of numbers [#permalink]

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New post 11 Mar 2015, 08:13
Nihit wrote:
A certain computer program generates a sequence of numbers a1, a2, … , an such that a1 = a2 = 1 and ak = ak-1 + 2ak-2 for all integers k such that 3 ≤ k ≤ n. If n > 6, then a7 = ?

A. 32
B. 43
C. 64
D. 100
E. 128


Hi,

It will help if (k-1) and (k-2) are written as subscript. Initially, I understood it to be (ak-1+2ak-2) which, without sub-scripting equals ak + 2ak - 3.

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Re: A certain computer program generates a sequence of numbers [#permalink]

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Re: A certain computer program generates a sequence of numbers   [#permalink] 06 Oct 2017, 21:25
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