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# For all n such that n is a positive integer, the terms of a certain

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Math Expert
Joined: 02 Sep 2009
Posts: 43381
For all n such that n is a positive integer, the terms of a certain [#permalink]

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08 Jun 2015, 08:08
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For all n such that n is a positive integer, the terms of a certain sequence B are given by the following rules:

$$B_n = B_{n-1} + 5$$ if n is odd and greater than 1;
$$Bn = -B_{n-1}$$ if n is even;
$$B_1 = 3$$

What is the sum of the first 65 terms in the sequence?

(A) –5
(B) 0
(C) 3
(D) 5
(E) 8

Kudos for a correct solution.
[Reveal] Spoiler: OA

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Re: For all n such that n is a positive integer, the terms of a certain [#permalink]

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08 Jun 2015, 08:39
5
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Bunuel wrote:
For all n such that n is a positive integer, the terms of a certain sequence B are given by the following rules:

$$B_n = B_{n-1} + 5$$ if n is odd and greater than 1;
$$Bn = -B_{n-1}$$ if n is even;
$$B_1 = 3$$

What is the sum of the first 65 terms in the sequence?

(A) –5
(B) 0
(C) 3
(D) 5
(E) 8

Kudos for a correct solution.

B1 = 3
B2 = -3
B3 = 2
B4 = -2
B5 = 3 and from this moment we see that pattern will be repeat each four numbers
65 / 4 = 4*16 + 1

so B65 will be equal to 3 (16 repetions + 1 more calculation)

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Joined: 27 Dec 2013
Posts: 299
Re: For all n such that n is a positive integer, the terms of a certain [#permalink]

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08 Jun 2015, 08:55
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The first four terms in the sequence are 3, -3,+3,-3. Repeat teh sequence for 16 times. You get 64 numbers and the sum is 0. The last number will be 3.

KR.

Bunuel wrote:
For all n such that n is a positive integer, the terms of a certain sequence B are given by the following rules:

$$B_n = B_{n-1} + 5$$ if n is odd and greater than 1;
$$Bn = -B_{n-1}$$ if n is even;
$$B_1 = 3$$

What is the sum of the first 65 terms in the sequence?

(A) –5
(B) 0
(C) 3
(D) 5
(E) 8

Kudos for a correct solution.

_________________

Kudos to you, for helping me with some KUDOS.

Manager
Joined: 26 Dec 2011
Posts: 120
Schools: HBS '18, IIMA
Re: For all n such that n is a positive integer, the terms of a certain [#permalink]

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08 Jun 2015, 09:10
1
KUDOS
First four terms are repeating B1=3, B2=-3, B3=2, B4=-2....

There are 16 such repetitions equal to 64 and the 65th term is B65=B1=3.

Thanks
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Manager
Joined: 21 Jan 2015
Posts: 149
Location: India
Concentration: Strategy, Marketing
WE: Marketing (Other)
Re: For all n such that n is a positive integer, the terms of a certain [#permalink]

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09 Jun 2015, 01:00
1
KUDOS
Bunuel wrote:
For all n such that n is a positive integer, the terms of a certain sequence B are given by the following rules:

$$B_n = B_{n-1} + 5$$ if n is odd and greater than 1;
$$Bn = -B_{n-1}$$ if n is even;
$$B_1 = 3$$

What is the sum of the first 65 terms in the sequence?

(A) –5
(B) 0
(C) 3
(D) 5
(E) 8

Ans: C

Solution: by given condition we start from B1=3, B2=-3, B3=2, B4=-2, B5=3, B6=-3 B7=2, B8=-2....... and so on
so this sequence repeat first 4 terms with the cycle of 4 terms. and there sum is 0.
so now first 65 terms had 16 full cycle and one extra term B65=3

so the sum of first 64 terms is 3.
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Re: For all n such that n is a positive integer, the terms of a certain [#permalink]

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10 Jun 2015, 00:56
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Bn=Bn−1+5 if n is odd and greater than 1;
Bn=−Bn−1 if n is even;
B1=3
$$B_1 =3$$
$$B_2 =-B_1=-3$$ because 2 is even
$$B_3 =B_2+5=2$$
$$B_4 =-B_3=-2$$
$$B_5 =B_4+5=3$$
$$B_6 =-B_5=-3$$ etc.
so it is $$3-3+2-2+3-3+...$$ 35 times, with every two numbers summing to 0. For the first 65 terms, there are 32 pairs and then the 65th term. The 65th term is 3, because it starts the 33th pair, and the start of every odd pair is 3. So the answer is C
Manager
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Posts: 56
Re: For all n such that n is a positive integer, the terms of a certain [#permalink]

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14 Jun 2015, 03:47
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KUDOS
The first four terms are 3, -3,+3,-3.
sequence for 64 numbers = sum is 0.
Last number will be 3.

Math Expert
Joined: 02 Sep 2009
Posts: 43381
Re: For all n such that n is a positive integer, the terms of a certain [#permalink]

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15 Jun 2015, 02:45
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Bunuel wrote:
For all n such that n is a positive integer, the terms of a certain sequence B are given by the following rules:

$$B_n = B_{n-1} + 5$$ if n is odd and greater than 1;
$$Bn = -B_{n-1}$$ if n is even;
$$B_1 = 3$$

What is the sum of the first 65 terms in the sequence?

(A) –5
(B) 0
(C) 3
(D) 5
(E) 8

Kudos for a correct solution.

MANHATTAN GMAT OFFICIAL SOLUTION:

Let's list the first few terms of the sequence according to the given rules:
B1 = 3
B2 = -B1 = –3
B3 = B2 + 5 = –3 + 5 = 2
B4 = -B3 = –2
B5 = B4 + 5 = –2 + 5 = 3
…etc.

Note that the pattern is a four-term repeat: 3, –3 , 2, –2 . Also note that the sum of this repeating group is (3) + (–3) + (2) + (–2) = 0. This repeating group will occur 16 times through term number 64. Thus, the sum of the first 64 terms will be 0. This leaves the 65th term, which will have the same value as B1 : 3. Therefore, the sum of the first 65 terms is 3.

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Re: For all n such that n is a positive integer, the terms of a certain [#permalink]

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04 Aug 2017, 12:54
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Re: For all n such that n is a positive integer, the terms of a certain   [#permalink] 04 Aug 2017, 12:54
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