Manhattan GMAT Challenge Problem of the Week - 10 May 2011

by Manhattan Prep, May 10, 2011

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Question

For all positive integers n, the nth term in sequence [pmath]S_n[/pmath] is defined as follows:

[pmath]S_n[/pmath] = [pmath](n!)^{-1}[/pmath]

The sum of the first six terms of [pmath]S_n[/pmath] is

(A) between 0 and

(B) between and 1

(C) between 1 and 1

(D) between 1 and 2

(E) greater than 2

Answer

First, list out the first few terms of [pmath]S_n[/pmath] to get a feeling for the sequence.

[pmath]S_1[/pmath] = [pmath](1!)^{-1}[/pmath] = 1/1! = 1/1 = 1

[pmath]S_2[/pmath] = [pmath](2!)^{-1}[/pmath] = 1/2! = 1/2 =

The sum of just the first two terms, by the way, is 1. Glancing ahead at the answers, we should already be able to rule out A, B, and C at this point. The terms never go negative, so the only question is whether the sum of the first six terms takes us above 2 or not.

The sum of the first six terms can be written this way:

Sum = 1/1! + 1/2! + 1/3! + 1/4! + 1/5! + 1/6!

= 1 + 1/2 + 1/6 + 1/24 + 1/120 + 1/720

Given how quickly the fractions shrink, we might guess at this point that the sum does not go above 2. However, we can make sure of this guess. Leaving out the first two terms (the 1 and the ), we can just figure out the sum of the last four terms:

1/6 + 1/24 + 1/120 + 1/720

= 1/6(1 + 1/4 + 1/20 + 1/120)

= 1/6(120/120 + 30/120 + 6/120 + 1/120)

= 1/6(157/120)

= 157/720,

which is definitely less than 1/2. (If it were above 1/2, then the sum of the first six terms would be greater than 2.)

Thus, the sum of the first six terms is between 1 and 2.

The correct answer is D.

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