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What of the followings is the range of X = 1 + (1/2^2) + (1/3^2)

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What of the followings is the range of X = 1 + (1/2^2) + (1/3^2) [#permalink]

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New post 06 Nov 2017, 21:52
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[GMAT math practice question]

What of the followings is the range of X = 1 + (1/2^2) + (1/3^2) + ...... + (1/100^2),

A. 0 < X < 1
B. 1 < X < 2
C. 2 < X < 3
D. 3 < X < 4
E. 4 < X < 5

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What of the followings is the range of X = 1 + (1/2^2) + (1/3^2) [#permalink]

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New post 07 Nov 2017, 09:18
1
MathRevolution wrote:
[GMAT math practice question]

What of the followings is the range of X = 1 + (1/2^2) + (1/3^2) + ...... + (1/100^2),

A. 0 < X < 1
B. 1 < X < 2
C. 2 < X < 3
D. 3 < X < 4
E. 4 < X < 5


clearly \(x>1\) as it is a positive summation over \(1\). the challenge is to find the maximum value of \(\frac{1}{2^2} + \frac{1}{3^2} +.......+\frac{1}{{100}^2}\)

Now, \(\frac{1}{2^2} + \frac{1}{3^2} +.......+\frac{1}{{100}^2} < \frac{1}{1*2} + \frac{1}{2*3} + \frac{1}{3*4}................+\frac{1}{99*100}\)

or, \(\frac{1}{2^2} + \frac{1}{3^2} +.......+\frac{1}{{100}^2} < 1-\frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4}................+\frac{1}{99} - \frac{1}{100}\)

or, \(\frac{1}{2^2} + \frac{1}{3^2} +.......+\frac{1}{{100}^2} < 1-\frac{1}{100}\), now clearly \(1-\frac{1}{100}<1\)

so \(\frac{1}{2^2} + \frac{1}{3^2} +.......+\frac{1}{{100}^2} < 1\). Add \(1\) to both sides of the inequality

Hence \(1+ \frac{1}{2^2} + \frac{1}{3^2} +.......+\frac{1}{{100}^2} < 1+1\)

or \(1<x<2\)

Option B
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What of the followings is the range of X = 1 + (1/2^2) + (1/3^2) [#permalink]

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New post 09 Nov 2017, 00:56
=>

\(\frac{1}{A^2} < \frac{1}{(A-1)A} = \frac{1}{A-1} – \frac{1}{A}\)

It is clear that \(1 + \frac{1}{2^2} + \frac{1}{3^2} + ...... + \frac{1}{100^2} > 1\).

\(1 + \frac{1}{2^2} + \frac{1}{3^2} + ...... + \frac{1}{100^2}\)
\(< 1 + \frac{1}{{1 \cdot 2}} + \frac{1}{{2 \cdot 3}} +\frac{1}{{3 \cdot 4}} + … + \frac{1}{{99 \cdot 100}}\)
\(= 1 + ( \frac{1}{1} – \frac{1}{2} ) + ( \frac{1}{2} – \frac{1}{3} ) + … + (\frac{1}{99} – \frac{1}{100})\)
\(= 1 + \frac{1}{1} – \frac{1}{100} = 2 – \frac{1}{100} < 2\)

Thus \(1 < X < 2\).
Therefore, B is the answer.

Answer: B
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Re: What of the followings is the range of X = 1 + (1/2^2) + (1/3^2) [#permalink]

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New post 14 Jun 2018, 05:27
MathRevolution wrote:
=>

\(\frac{1}{A^2} < \frac{1}{(A-1)A} = \frac{1}{A-1} – \frac{1}{A}\)

It is clear that \(1 + \frac{1}{2^2} + \frac{1}{3^2} + ...... + \frac{1}{100^2} > 1\).

\(1 + \frac{1}{2^2} + \frac{1}{3^2} + ...... + \frac{1}{100^2}\)
\(< 1 + \frac{1}{{1 \cdot 2}} + \frac{1}{{2 \cdot 3}} +\frac{1}{{3 \cdot 4}} + … + \frac{1}{{99 \cdot 100}}\)
\(= 1 + ( \frac{1}{1} – \frac{1}{2} ) + ( \frac{1}{2} – \frac{1}{3} ) + … + (\frac{1}{99} – \frac{1}{100})\)
\(= 1 + \frac{1}{1} – \frac{1}{100} = 2 – \frac{1}{100} < 2\)

Thus \(1 < X < 2\).
Therefore, B is the answer.

Answer: B



Hi MathRevolution,

The below step is not clear , can you please explain

\(< 1 + \frac{1}{{1 \cdot 2}} + \frac{1}{{2 \cdot 3}} +\frac{1}{{3 \cdot 4}} + … + \frac{1}{{99 \cdot 100}}\)

Thank you.
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Re: What of the followings is the range of X = 1 + (1/2^2) + (1/3^2)   [#permalink] 14 Jun 2018, 05:27
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