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

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New post 09 Sep 2015, 00:58
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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


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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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New post 09 Sep 2015, 02:11
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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
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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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New post 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.

The correct answer is B.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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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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New post 18 Apr 2017, 10:18
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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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New post 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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