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S is a sequence such that Sn = (–1)n for each integer n ≥ 1.
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30 Jul 2018, 10:11
30 Jul 2018, 10:11
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97% (00:33) correct
2% (00:33) wrong based on 49 sessions
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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?
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Re: S is a sequence such that Sn = (–1)n for each integer n ≥ 1.
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11 Aug 2018, 12:37
11 Aug 2018, 12:37
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(-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.
-1+1-1+1...-1+1
Which yields 0.
Re: S is a sequence such that Sn = (–1)n for each integer n ≥ 1.
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12 Aug 2018, 06:58
12 Aug 2018, 06:58
Expert Reply
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.
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.
