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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2960
Given Kudos: 1290
GPA: 3.39
Sequence S is defined as a_n = (-1)^n(a_n – 1 + a_n – 2), where a_1 =
[#permalink]
09 Jul 2021, 08:38
09 Jul 2021, 08:38
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Question Stats:
47% (03:39) correct
52% (02:53) wrong based on 34 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
(A) -1
(B) 0
(C) 1
(D) 1597
(E) 2584
Re: Sequence S is defined as a_n = (-1)^n(a_n – 1 + a_n – 2), where a_1 =
[#permalink]
11 Jul 2021, 05:57
11 Jul 2021, 05:57
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According to the formula,
a_35 = - a_34 - a_33;
and
a_38 = a_37 + a_36;
Therefore,
a_35 + a_38 = a_35 + a_36 + a_37;
Notice that the required expression results in the sum of three consecutive terms: a_35 + a_36 + a_37;
If n is odd, then in the given formula reduces to:
a_n = - a_{n-1} - a_{n-2};
a_{n-2} + a_{n-1} + a_n = 0;
Therefore, if n is odd, then the sum of three consecutive terms in this sequence is 0.
In a_35 + a_36 + a_37;, n is 35 which is odd,
Therefore, a_35 + a_36 + a_37 = 0;
Therefore, the answer is B.
a_35 = - a_34 - a_33;
and
a_38 = a_37 + a_36;
Therefore,
a_35 + a_38 = a_35 + a_36 + a_37;
Notice that the required expression results in the sum of three consecutive terms: a_35 + a_36 + a_37;
If n is odd, then in the given formula reduces to:
a_n = - a_{n-1} - a_{n-2};
a_{n-2} + a_{n-1} + a_n = 0;
Therefore, if n is odd, then the sum of three consecutive terms in this sequence is 0.
In a_35 + a_36 + a_37;, n is 35 which is odd,
Therefore, a_35 + a_36 + a_37 = 0;
Therefore, the answer is B.
General Discussion
Re: Sequence S is defined as a_n = (-1)^n(a_n 1 + a_n 2), where a_1 =
[#permalink]
28 Oct 2025, 21:43
28 Oct 2025, 21:43
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Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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