Sequence S is defined as a_n = (-1)^n(a_n 1 + a_n 2), where a_1 =

MANHATTAN GMAT OFFICIAL SOLUTION:

The first step in solving this problem is to figure out the first several terms of S, after the first two (which equal 1):

\(a_3 = (-1)^3(a_2 + a_1) = (-1)(1 + 1) = -2\)
\(a_4 = (-1)^4(a_3 + a_2) = (1)(-2 + 1) = -1\)
\(a_5 = (-1)^5(a_4 + a_3) = (-1)(-1 + -2) = 3\)
\(a_6 = (-1)^6(a_5 + a_4) = (1)(3 + -1) = 2\)
\(a_7 = (-1)^7(a_6 + a_5) = (-1)(3 + 2) = -5\)
\(a_8 = (-1)^8(a_7 + a_8) = (1)(-5 + 2) = -3\)

Unfortunately, the terms do not seem to be repeating, but we might notice that each even term is equal to the negative of the term 3 positions prior.

This would mean that \(a_35 + a_38\) would equal zero.

Let’s test this hypothesis by working backwards from the expression. Note that each even term has a positive 1 in front as a factor, whereas each odd term has -1 as a factor.

\(a_35 + a_38\)
\(= a_35 + (a_37 + a_36)\)
\(= a_35 + ((-a_36 – a_35)+ a_36)\)
= 0

Incidentally, the name of the problem is inspired by the famous Fibonacci sequence, in which each term after the first two is the sum of the two previous terms (and terms #1 and #2 are both equal to 1): 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, etc.

The alternating sign in the definition of our sequence throws a monkey wrench into the works, but in many respects the sequence in this problem is Fibonacci-like: all of the terms have values equal to Fibonacci numbers or the negatives of Fibonacci numbers.

The correct answer is B.