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If S = 1 + 1/2^2 + 1/3^2 + 1/4^2 + 1/5^2 + 1/6^2 + 1/7^2 + 1/8^2 + 1/9

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If S = 1 + 1/2^2 + 1/3^2 + 1/4^2 + 1/5^2 + 1/6^2 + 1/7^2 + 1/8^2 + 1/9  [#permalink]

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New post 25 Jun 2018, 05:28
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Question Stats:

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If \(S = 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} + \frac{1}{7^2} + \frac{1}{8^2} + \frac{1}{9^2} + \frac{1}{10^2}\), which of the following is true?


A. S > 3

B. S = 3

C. 2 < S < 3

D. S = 2

E. S < 2



NEW question from GMAT® Official Guide 2019


(PS06243)

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Re: If S = 1 + 1/2^2 + 1/3^2 + 1/4^2 + 1/5^2 + 1/6^2 + 1/7^2 + 1/8^2 + 1/9  [#permalink]

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New post 25 Jun 2018, 06:03
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Bunuel wrote:
If \(S = 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} + \frac{1}{7^2} + \frac{1}{8^2} + \frac{1}{9^2} + \frac{1}{10^2}\), which of the following is true?


A. S > 3

B. S = 3

C. 2 < S < 3

D. S = 2

E. S < 2




\(S = 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} + \frac{1}{7^2} + \frac{1}{8^2} + \frac{1}{9^2} + \frac{1}{10^2}\) = 1 + 1/4 + 1/9 + 1/16 + 1/25 + 1/36 + . . . .

Now, let's convert a few of the fractions to decimal approximations...
S = 1 + 0.25 + 0.11 + 0.06 + 0.04 + . . .

Add the first 4 values....
S = 1.42 + 0.04 + . . .

IMPORTANT: Notice that each fraction is less than the fraction before it
So, all of the 5 decimals after 0.04 will be less than 0.04
So, if we replace all of those 5 decimals with 0.04, our new sum will be greater than the original sum

So: S < 1.42 + 0.04 + 0.04 + 0.04 + 0.04 + 0.04 + 0.04

Simplify: S < 1.42 + 0.24
Simplify: S < 1.66

Answer: E

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Brent
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If S = 1 + 1/2^2 + 1/3^2 + 1/4^2 + 1/5^2 + 1/6^2 + 1/7^2 + 1/8^2 + 1/9  [#permalink]

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New post 25 Jun 2018, 06:05
2
Bunuel wrote:
If \(S = 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} + \frac{1}{7^2} + \frac{1}{8^2} + \frac{1}{9^2} + \frac{1}{10^2}\), which of the following is true?


A. S > 3

B. S = 3

C. 2 < S < 3

D. S = 2

E. S < 2



NEW question from GMAT® Official Guide 2019


(PS06243)

Hi,

For simplicity, let's break the given series in three parts as follows:

\(S_{1} = 1\),

\(S_{2} = \frac{1}{2^2} + \frac{1}{3^2}\), and

\(S_{3} = \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} + \frac{1}{7^2} + \frac{1}{8^2} + \frac{1}{9^2} + \frac{1}{10^2}\)

In \(S_{2}\) larger term is \(\frac{1}{2^2}\), and there are two terms. Hence,

\(S_{2} = \frac{1}{2^2} + \frac{1}{3^2} < 2*\frac{1}{4} = \frac{1}{2}\) --- (1)

Similarly, in series \(S_{3}\) the largest term is \(\frac{1}{4^2} = \frac{1}{16}\) and there are total 7 terms. Hence,

\(S_{3} = \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} + \frac{1}{7^2} + \frac{1}{8^2} + \frac{1}{9^2} + \frac{1}{10^2} < 7* \frac{1}{16} < 8*\frac{1}{16} = \frac{1}{2}\) --- (2)

\(S = S_{1} + S_{2} + S_{3} < 1 + \frac{1}{2} + \frac{1}{2} = 2\)

Hence \(S < 2\). Answer (E).

Thanks.
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Re: If S = 1 + 1/2^2 + 1/3^2 + 1/4^2 + 1/5^2 + 1/6^2 + 1/7^2 + 1/8^2 + 1/9  [#permalink]

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New post 25 Jun 2018, 06:07
Since the terms after 1/9 add on to the hundreths place and since its easy to calculate upto 1/25, let's maximize all the terms after 1/25 by assuming the remaining terms as 1/25.

S = 1 + 0.25 + 0.11 + 0.0625 + 0.04*6
S = 1.4225 + 0.24 = 1.6625 < 2

Answer: E
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Re: If S = 1 + 1/2^2 + 1/3^2 + 1/4^2 + 1/5^2 + 1/6^2 + 1/7^2 + 1/8^2 + 1/9  [#permalink]

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New post 10 Jul 2018, 13:26
Can smbd post explanation from OG?
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If S = 1 + 1/2^2 + 1/3^2 + 1/4^2 + 1/5^2 + 1/6^2 + 1/7^2 + 1/8^2 + 1/9  [#permalink]

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New post 10 Jul 2018, 23:59
[quote="Bunuel"]If \(S = 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} + \frac{1}{7^2} + \frac{1}{8^2} + \frac{1}{9^2} + \frac{1}{10^2}\), which of the following is true?


A. S > 3

B. S = 3

C. 2 < S < 3

D. S = 2

E. S < 2

We have the terms after \(\frac{1}{5^2}\) are close to zero. Hence the sum of the terms henceforth \(\frac{1}{5^2}\) can be approximated to zero.
Hence \(S = 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} + \frac{1}{7^2} + \frac{1}{8^2} + \frac{1}{9^2} + \frac{1}{10^2}\) reduces to
\(S=1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2}\)
Or, \(S=\frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25}\)=1+0.25+0.11+0.0625+0.04=1.4625<2
So S<2.

Ans (E)
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Re: If S = 1 + 1/2^2 + 1/3^2 + 1/4^2 + 1/5^2 + 1/6^2 + 1/7^2 + 1/8^2 + 1/9  [#permalink]

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New post 07 Feb 2019, 05:14
I tried to solve it like below:
S= 1+ 1/2^2 + 1/3^2 + 1/4^2 + ...
S= 1 + {(1/6^2) * 9} [considering 1/6^2 as the the middle term in the remaining 9 terms. And multiplying it by number of terms. To get an approx value]
S= 1 + 1/4
S= 1.25
Hence (E)

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If S = 1 + 1/2^2 + 1/3^2 + 1/4^2 + 1/5^2 + 1/6^2 + 1/7^2 + 1/8^2 + 1/9  [#permalink]

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New post 07 Feb 2019, 06:18
\(S = 1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} + \frac{1}{7^2} + \frac{1}{8^2} + \frac{1}{9^2} + \frac{1}{10^2}\)

GMAT quant questions tend to be designed to be answered efficiently via the use of hacks. So, let's start hacking.

Start off with the obvious: \(1 + \frac{1}{2^2} = 1 \frac{1}{4}\)

It becomes apparent that to get to 2 we need 3/4 more, and those other fractions look pretty small. They probably won't get us to 2.

If we can prove that 1 1/4 + (the sum of the rest of those fractions) < 2, then the correct answer will be S < 2 and we'll be done.

The next fraction is \(\frac{1}{3^2}\).

That is effectively \(\frac{1}{9}\).

We have 1 1/4 + 1/9.

Without adding, we know that 1 1/4 + 1/9 < 1 1/2.

The next seven are \(\frac{1}{4^2} + \frac{1}{5^2} + \frac{1}{6^2} + \frac{1}{7^2} + \frac{1}{8^2} + \frac{1}{9^2} + \frac{1}{10^2}\).

Since 8 x 1/16 = 1/2, and the first fraction of this set of seven fractions is 1/16 and all the rest are smaller, this set has to add up to less than 1/2.

So, S = 1 1/4 + 1/9 + (a number < 1/2).

S < 2.

The correct answer is (E).
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If S = 1 + 1/2^2 + 1/3^2 + 1/4^2 + 1/5^2 + 1/6^2 + 1/7^2 + 1/8^2 + 1/9   [#permalink] 07 Feb 2019, 06:18
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