Manhattan GMAT Challenge Problem of the Week – 28 Mar 2011

by Manhattan Prep, Mar 28, 2011

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Question

The first term in sequence Q equals 1, and for all positive integers n equal to or greater than 2, the nth term in sequence Q equals the absolute value of the difference between the nth smallest positive perfect cube and the (n-1)st smallest positive perfect cube. The sum of the first seven terms in sequence Q is

(A) 91

(B) 127

(C) 216

(D) 343

(E) 784

Answer

To solve this problem, we must translate the verbal instructions for the construction of the sequence. The first term is easy: [pmath]Q_1[/pmath] = 1.

Next, we have this difficult wording: for all positive integers n equal to or greater than 2, the nth term in sequence Q equals the absolute value of the difference between the nth smallest positive perfect cube and the (n-1)st smallest positive perfect cube.

So were dealing with all the positive integers beyond 1. Lets take as an example n = 2. The instructions become these: the second term equals the absolute value of the difference between the second (nth) smallest positive perfect cube and the first (that is, n-1st) smallest positive perfect cube.

We know we need to consider the positive perfect cubes in order:

[pmath]1^3[/pmath] = 1 = smallest positive perfect cube (or first smallest).

[pmath]2^3[/pmath] = 8 = second smallest positive perfect cube.

The absolute value of the difference between these cubes is 8 1 = 7. Thus [pmath]Q_2[/pmath] = 8 1 = 7.

Likewise, [pmath]Q_3[/pmath] = |[pmath]3^3[/pmath] [pmath]2^3[/pmath]| = 27 8 = 19, and so on.

Now, rather than figure out each term of Q separately, then add up, we can save time if we notice that the cumulative sums telescope in a simple way. This is what telescoping means:

The sum of [pmath]Q_2[/pmath] and [pmath]Q_1[/pmath] = 7 + 1 = 8. We can also write (8 - 1) + 1 = 8. Notice how the 1s cancel.

The sum of [pmath]Q_3[/pmath], [pmath]Q_2[/pmath] and [pmath]Q_1[/pmath] = 19 + 7 + 1 = 27. We can also write (27 - 8 ) + (8 - 1) + 1 = 27. Notice how the 8s and the 1s cancel.

At this point, we hopefully notice that the cumulative sum of [pmath]Q_1[/pmath] through [pmath]Q_n[/pmath] is just the nth smallest positive perfect cube.

So the sum of the first seven terms of the sequence is [pmath]7^3[/pmath], which equals 343.

The correct answer is D.

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