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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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)

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.
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One point to recall here is that zero is neither positive or negative, so we don't need to use it in the set of n numbers.

How do you know the sequence goes like that? nth value is 1, then 2,then 3 and so forth? Is it from how the problem is worded? "for each positive integer n?"

Yes.

You from zero up on the number line

Each positive integer.............
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Don't you have to add +1 at the end of the nth of terms. I found 19 even integers and then I added +1 to make the result 40. What did I do wrong?

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