Notice that the 10th and 11th terms (the two middle terms in a set of 20 terms) are arithmetic inverses, that is, their sum is zero. Likewise, the 1st and 20th terms sum to zero, as do the 2nd and 19th terms. In the first 20 terms of the sequence, we can make 10 pairs that each sum to zero. Thus, Quantity A is zero.

For the sum of the first 19 terms, we could either

(1) Subtract a_{20}from the sum of the first 20 terms: 0 – (2^{20}– 2^{1}) = 2^{1}– 2^{20}= 2 – a very large positive number = negative, or

(2) Realize that in the first 19 terms, all terms except a_{1}can be paired such that the pair sums to zero, so the sum of the first 19 terms = a_{1}= 2^{1}– 2^{20}= 2 – a very large positive number = negative.

Thus, Quantity B is negative, which is less than zero.

The correct answer is A.

This isn't far removed from what I said in this comment.

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