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.
_________________

Sandy
If you found this post useful, please let me know by pressing the Kudos Button

Try our free Online GRE Test

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.