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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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Nihit
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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Nihit
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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43...choice B

a5=11; a6=21; a7=43
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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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even though this is good observation. this approach wont work if there are more than 1 odd number in the answer choices.

chix475ntu
Nihit
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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Nihit
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
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x2suresh
Nihit
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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rishi2377
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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rishi2377
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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calculating in sequence

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

Ans B
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Nihit
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.

TO
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