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The sequence of numbers a1, a2, a3, …, an, … is defined by

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The sequence of numbers a1, a2, a3, …, an, … is defined by [#permalink] New post 16 Sep 2017, 11:06
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The sequence of numbers a1, a2, a3, …, an, … is defined by
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for each integer n ≥ 1

Quantity A
Quantity B
The sum of the first 32 terms of this sequence
The sum of the first 31 terms of this sequence


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

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Re: The sequence of numbers a1, a2, a3, …, an, … is defined by [#permalink] New post 21 Sep 2017, 07:55
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Since until the 31st term the two sums are equal we can focus on the 32nd term, which is equal to \(2^{32}-\frac{1}{2^{32-33}}=2^{32}-\frac{1}{2^{-1}}=2^{32}-2>0\). Thus quantity A is larger!
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Re: The sequence of numbers a1, a2, a3, …, an, … is defined by [#permalink] New post 22 Feb 2018, 06:48
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With a little concern, we can find out the answer. Since 31st terms of both sides are equal, we can eliminate them. Then by calculating, we can see that the quantity A is bigger than the quantity B.
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Re: The sequence of numbers a1, a2, a3, …, an, … is defined by [#permalink] New post 22 Feb 2018, 10:14
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As IlCreatore said, notice that Quantity A and Quantity B have a lot of overlap. They both contain the first 31 terms of the sequences, you we can subtract that from both sides. We would thus be left with just the 32nd term under Quantity A and 0 under Quantity B. So the real question is whether the 32nd term is greater than 0.

Since \(\frac{1}{2^{32-33}}=\frac{1}{2^{-1}}=2\), and since \(2^{32}\) is some gigantic number, we know that a gigantic number minus 2 is larger than 0, so the answer is A.
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Re: The sequence of numbers a1, a2, a3, …, an, … is defined by [#permalink] New post 24 Feb 2018, 17:22
A
Re: The sequence of numbers a1, a2, a3, …, an, … is defined by   [#permalink] 24 Feb 2018, 17:22
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