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# The infinite sequence a(1), a(2),, a(n), is such that a(1) =

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VP
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The infinite sequence a(1), a(2),, a(n), is such that a(1) = [#permalink]

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10 Dec 2007, 15:08
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The infinite sequence a(1), a(2),…, a(n),… is such that a(1) = 2, a(2) = -3, a(3) = 5, a(4) = -1, and a(n) = a(n-4) for n > 4. What is the sum of the first 97 terms of the sequence?

A. 72
B. 74
C. 75
D. 78
E. 80

B
CEO
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10 Dec 2007, 15:15
B.

a(97)=a(4*24+1)

∑a(1..97)=24*∑a(1..4)+a(97)=24*3+a(1)=72+2=74
VP
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10 Dec 2007, 19:07
thanks walker. no other simpler methods??
Director
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10 Dec 2007, 19:33
walker wrote:
B.

a(97)=a(4*24+1)

∑a(1..97)=24*∑a(1..4)+a(97)=24*3+a(1)=72+2=74

can you break this down step by step for me? thanks in advance
Director
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11 Dec 2007, 03:44
Ravshonbek wrote:
The infinite sequence a(1), a(2),…, a(n),… is such that a(1) = 2, a(2) = -3, a(3) = 5, a(4) = -1, and a(n) = a(n-4) for n > 4. What is the sum of the first 97 terms of the sequence?

A. 72
B. 74
C. 75
D. 78
E. 80

B

74.

a(5) = a(4)
a(6) = a(2) and so on.

So the series is basically the first 4 terms repeated 97 times:
2, -3 , 5, -1, 2, -3 , 5, -1, ... and so on.

Sum of the 1st 4 terms = 3
These same terms are repeated 97 = 96 + 1 times
Therefore, the sum will be 96/4*3 + 1st term = 72 + 2 = 74
CEO
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11 Dec 2007, 03:55
eschn3am wrote:
can you break this down step by step for me? thanks in advance

1 a(n) = a(n-4) for n > 4: we have the sequence of numbers with period 4

2. for any M=4*k+r: a(M)=a(r)

3. we can transform ∑a(1..M) to:

∑a(1..M) = ∑a(1..4) + ∑a(5..8) + .... + ∑a(4*(k-1)+1..4*k)... + ∑a(1..r)

put ∑a(1..4) insead of ∑a(5..8) and ∑a(4*(k-1)+1..4*k))

∑a(1..M) = k*∑a(1..4)+ ∑a(1..r)

4. M=97=4*24+1 ==> ∑a(1..97) = 24*∑a(1..4)+ ∑a(1..1) = 24*3+a(1)=72+2=74
11 Dec 2007, 03:55
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