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# S is a sequence such that Sn = (–1)n for each integer n ≥ 1.

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GMAT Club Legend
Joined: 07 Jun 2014
Posts: 4749
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
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Kudos [?]: 1659 [0], given: 396

S is a sequence such that Sn = (–1)n for each integer n ≥ 1. [#permalink]  30 Jul 2018, 10:11
Expert's post
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Question Stats:

100% (00:18) correct 0% (00:00) wrong based on 5 sessions
S is a sequence such that $$S_n = (-1)^n$$ for each integer n ≥ 1. What is the sum of the first 20 terms in S?

[Reveal] Spoiler: OA
0

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Sandy
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Intern
Joined: 10 Aug 2018
Posts: 33
Followers: 0

Kudos [?]: 12 [0], given: 0

Re: S is a sequence such that Sn = (–1)n for each integer n ≥ 1. [#permalink]  11 Aug 2018, 12:37
(-1)^n equals 1 if n is even, and -1 if n is odd. There is an equal number of odd and even numbers between 1 and 20, so the sum will look like this:

-1+1-1+1...-1+1

Which yields 0.
GMAT Club Legend
Joined: 07 Jun 2014
Posts: 4749
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
Followers: 93

Kudos [?]: 1659 [0], given: 396

Re: S is a sequence such that Sn = (–1)n for each integer n ≥ 1. [#permalink]  12 Aug 2018, 06:58
Expert's post
Explanation

Adding 20 individual terms would take quite a long time. Look for a pattern. The first several terms in $$S_n = (-1)^n$$, where n ≥ 1:

$$S_1 = (-1)1 = -1$$

$$S_2 = (-1)2 = 1$$

$$S_3 = (-1)3 = -1$$

$$S_4 = (-1)4 = 1$$

The terms alternate –1, 1, –1, 1, and so on. If the terms are added, every pair of –1 and 1 will add to zero; in other words, for an even number of terms, the sum will be zero. Since 20 is an even number, so the first 20 terms sum to zero.
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Re: S is a sequence such that Sn = (–1)n for each integer n ≥ 1.   [#permalink] 12 Aug 2018, 06:58
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