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For each positive integer n, the nth term of the sequence S [#permalink]
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Expert's post 00:00

Question Stats: 79% (00:39) correct 20% (01:08) wrong based on 74 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.
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Re: For each positive integer n, the nth term of the sequence S [#permalink]
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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:

 1st 2nd 3rd 4th 5th 6th 7th ... 0 2 0 2 0 2 0 ...

How many even numbers from 1 to 39? 19

Thus the sum = $$19 \times 2 = 38$$

Quantity B is greater.
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Sandy
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Re: For each positive integer n, the nth term of the sequence S [#permalink]
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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:

 1st 2nd 3rd 4th 5th 6th 7th ... 0 2 0 2 0 2 0 ...

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) Manager Joined: 22 Feb 2018
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Re: For each positive integer n, the nth term of the sequence S [#permalink]
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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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