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For each positive integer n, the nth term of the sequence S

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For each positive integer n, the nth term of the sequence S [#permalink] New post 24 Feb 2017, 02:29
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82% (00:31) correct 17% (01:07) wrong based on 46 sessions


For each positive integer n, the \(N\)th term of the sequence S is \(1 +\) (\(-1^n\)).

Quantity A
Quantity B
The sum of the first 39 terms of S
39


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

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Re: For each positive integer n, the nth term of the sequence S [#permalink] New post 04 Mar 2017, 18:12
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Explanation

Start by determining the first few terms to see if there is a pattern, as there often is in sequence questions:

1st term: n = 1 \(1 + (-1)^n\) → \(1 + (-1)^1\) → 1 + –1 → 0
2nd term: n = 2 \(1 + (-1)^n\) → \(1 + (-1)^2\) → 1 + 1 → 2
3rd term: n = 3 \(1 + (-1)^n\) → \(1 + (-1)^3\) → 1 + –1 → 0
4th term: n = 4 \(1 + (-1)^n\) → \(1 + (-1)^4\) → 1 + 1 → 2

As you can see, all odd terms result in 0, while all even terms result in 2. So only the even terms will count in the sum:


1st2nd3rd4th5th6th7th ...
0202020...


How many even numbers from 1 to 39? 19

Thus the sum = \(19 \times 2 = 38\)

Quantity B is greater.
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Re: For each positive integer n, the nth term of the sequence S [#permalink] New post 27 Mar 2018, 23:53
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sandy wrote:
Explanation

Start by determining the first few terms to see if there is a pattern, as there often is in sequence questions:

1st term: n = 1 \(1 + (-1)^n\) → \(1 + (-1)^1\) → 1 + –1 → 0
2nd term: n = 2 \(1 + (-1)^n\) → \(1 + (-1)^2\) → 1 + 1 → 2
3rd term: n = 3 \(1 + (-1)^n\) → \(1 + (-1)^3\) → 1 + –1 → 0
4th term: n = 4 \(1 + (-1)^n\) → \(1 + (-1)^4\) → 1 + 1 → 2

As you can see, all odd terms result in 0, while all even terms result in 2. So only the even terms will count in the sum:


1st2nd3rd4th5th6th7th ...
0202020...


How many even numbers from 1 to 39? 19

Thus the sum = \(19 \times 2 = 38\)

Quantity B is greater.


I am not sure of the solution since in the stem, you place n in inside the parenthesis.

According to the the stem, each term would be 1+(-1)=0 (\(-1^2\)=-1 NOT 1)
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Re: For each positive integer n, the nth term of the sequence S [#permalink] New post 28 Mar 2018, 10:58
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Answer: B
Nth term is 1+ (-1)^n
Sum of the first 39 terms of S:
1+(-1)^1 + 1+ (-1)^2 + …. + 1+ (-1)^39 = 39 * 1 + [(-1)^1 + (-1)^2 + (-1)^3 + … + (-1)^39] = 39 *1 + [-1 + 1 + … + -1] = 39*1 -1 = 38
So B is bigger than A.
* there are 39 1s.
* all 1s and -1s nullify each other but the last one ((-1)^39) remains.
Re: For each positive integer n, the nth term of the sequence S   [#permalink] 28 Mar 2018, 10:58
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For each positive integer n, the nth term of the sequence S

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