Manhattan GMAT Challenge Problem of the Week – 22 Mar 2011

by Manhattan Prep, Mar 22, 2011

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Question

Sequence S is defined as [pmath]a_n[/pmath] = [pmath](-1)^n[/pmath]([pmath]a_n[/pmath] 1 + [pmath]a_n[/pmath] 2), where [pmath]a_1[/pmath] = 1 and [pmath]a_2[/pmath] = 1. What is the value of the sum [pmath]a_35[/pmath] + [pmath]a_38[/pmath] ?

(A) -1

(B) 0

(C) 1

(D) 1597

(E) 2584

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):

[pmath]a_3[/pmath] = [pmath](-1)^3[/pmath]([pmath]a_2[/pmath] + [pmath]a_1[/pmath]) = (-1)(1 + 1) = -2

[pmath]a_4[/pmath] = [pmath](-1)^4[/pmath]([pmath]a_3[/pmath] + [pmath]a_2[/pmath]) = (1)(-2 + 1) = -1

[pmath]a_5[/pmath] = [pmath](-1)^5[/pmath]([pmath]a_4[/pmath] + [pmath]a_3[/pmath]) = (-1)(-1 + -2) = 3

[pmath]a_6[/pmath] = [pmath](-1)^6[/pmath]([pmath]a_5[/pmath] + [pmath]a_4[/pmath]) = (1)(3 + -1) = 2

[pmath]a_7[/pmath] = [pmath](-1)^7[/pmath]([pmath]a_6[/pmath] + [pmath]a_5[/pmath]) = (-1)(3 + 2) = -5

[pmath]a_8[/pmath] = [pmath](-1)^8[/pmath]([pmath]a_7[/pmath] + [pmath]a_8[/pmath]) = (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 [pmath]a_35[/pmath] + [pmath]a_38[/pmath] 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.

[pmath]a_35[/pmath] + [pmath]a_38[/pmath]

= [pmath]a_35[/pmath] + ([pmath]a_37[/pmath] + [pmath]a_36[/pmath])

= [pmath]a_35[/pmath] + (([pmath]-a_36[/pmath] - [pmath]a_35[/pmath])+ [pmath]a_36[/pmath])

= 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.

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