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# Sequence S is defined as a_n = (-1)^n(a_n – 1 + a_n – 2), where a_1 =

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Math Expert
Joined: 02 Sep 2009
Posts: 46191
Sequence S is defined as a_n = (-1)^n(a_n – 1 + a_n – 2), where a_1 = [#permalink]

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09 Sep 2015, 00:58
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Difficulty:

65% (hard)

Question Stats:

68% (02:33) correct 32% (03:19) wrong based on 118 sessions

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Sequence S is defined as $$a_n = (-1)^n(a_{n - 1} + a_{n - 2})$$, where $$a_1 = 1$$ and $$a_2 = 1$$. What is the value of the sum $$a_{35} + a_{38}$$ ?

(A) -1
(B) 0
(C) 1
(D) 1597
(E) 2584

Kudos for a correct solution.

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Joined: 10 Aug 2015
Posts: 103
Re: Sequence S is defined as a_n = (-1)^n(a_n – 1 + a_n – 2), where a_1 = [#permalink]

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09 Sep 2015, 02:11
2
1
Bunuel wrote:
Sequence S is defined as $$a_n = (-1)^n(a_{n - 1} + a_{n - 2})$$, where $$a_1 = 1$$ and $$a_2 = 1$$. What is the value of the sum $$a_{35} + a_{38}$$ ?

(A) -1
(B) 0
(C) 1
(D) 1597
(E) 2584

Kudos for a correct solution.

Solution : $$a_{38} = (-1)^{38}(a_{38 - 1} + a_{38 - 2}) = a_{37} + a_{36} = (-1)^{37}(a_{37 - 1} + a_{37 - 2}) + a_{36} = - a_{36}- a_{35}+ a_{36} = - a_{35}$$
Therefore, $$a_{38} + a_{35} = 0$$

Option B
Math Expert
Joined: 02 Sep 2009
Posts: 46191
Re: Sequence S is defined as a_n = (-1)^n(a_n – 1 + a_n – 2), where a_1 = [#permalink]

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13 Sep 2015, 09:53
Bunuel wrote:
Sequence S is defined as $$a_n = (-1)^n(a_{n - 1} + a_{n - 2})$$, where $$a_1 = 1$$ and $$a_2 = 1$$. What is the value of the sum $$a_{35} + a_{38}$$ ?

(A) -1
(B) 0
(C) 1
(D) 1597
(E) 2584

Kudos for a correct solution.

MANHATTAN GMAT OFFICIAL SOLUTION:

The first step in solving this problem is to figure out the first several terms of S, after the first two (which equal 1):

$$a_3 = (-1)^3(a_2 + a_1) = (-1)(1 + 1) = -2$$
$$a_4 = (-1)^4(a_3 + a_2) = (1)(-2 + 1) = -1$$
$$a_5 = (-1)^5(a_4 + a_3) = (-1)(-1 + -2) = 3$$
$$a_6 = (-1)^6(a_5 + a_4) = (1)(3 + -1) = 2$$
$$a_7 = (-1)^7(a_6 + a_5) = (-1)(3 + 2) = -5$$
$$a_8 = (-1)^8(a_7 + a_8) = (1)(-5 + 2) = -3$$

Unfortunately, the terms do not seem to be repeating, but we might notice that each even term is equal to the negative of the term 3 positions prior.

This would mean that $$a_35 + a_38$$ would equal zero.

Let’s test this hypothesis by working backwards from the expression. Note that each even term has a positive 1 in front as a factor, whereas each odd term has -1 as a factor.

$$a_35 + a_38$$
$$= a_35 + (a_37 + a_36)$$
$$= a_35 + ((-a_36 – a_35)+ a_36)$$
= 0

Incidentally, the name of the problem is inspired by the famous Fibonacci sequence, in which each term after the first two is the sum of the two previous terms (and terms #1 and #2 are both equal to 1): 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, etc.

The alternating sign in the definition of our sequence throws a monkey wrench into the works, but in many respects the sequence in this problem is Fibonacci-like: all of the terms have values equal to Fibonacci numbers or the negatives of Fibonacci numbers.

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Joined: 18 Oct 2016
Posts: 139
Location: India
WE: Engineering (Energy and Utilities)
Re: Sequence S is defined as a_n = (-1)^n(a_n – 1 + a_n – 2), where a_1 = [#permalink]

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18 Apr 2017, 10:18
1
Option B

$$a38 = (-1)^{38}(a37 + a36) = (a37 + a36) = ((-1)^{37}(a36 + a35) + a36) = -a36 + (-a35) + a36 = -a35$$

$$a38 + a35 = 0$$
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Re: Sequence S is defined as a_n = (-1)^n(a_n – 1 + a_n – 2), where a_1 = [#permalink]

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23 May 2018, 14:47
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Re: Sequence S is defined as a_n = (-1)^n(a_n – 1 + a_n – 2), where a_1 =   [#permalink] 23 May 2018, 14:47
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