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.............

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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