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# The Quest for 700: Weekly GMAT Challenge (Answer)

Yesterday, Manhattan GMAT posted a GMAT question on our blog. Today, they have followed up with the answer:

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

a­­3 = (-1)3(a2 + a1) = (-1)(1 + 1) = -2

4 = (-1)4(a3 + a2) = (1)(-2 + 1) = -1

5 = (-1)5(a4 + a3) = (-1)(-1 + -2) = 3

6 = (-1)6(a5 + a4) = (1)(3 + -1) = 2

7 = (-1)7(a6 + a5) = (-1)(3 + 2) = -5

8 = (-1)8(a7 + a8) = (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 a35 + a38 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.

a35 + a38

a35 + (a37 + a36)

a35 + ((-a36a35)+ a36)

= 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.

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