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

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 6390
GMAT 1: 760 Q51 V42
GPA: 3.82
What of the followings is the range of X = 1 + (1/2^2) + (1/3^2)  [#permalink]

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06 Nov 2017, 21:52
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Difficulty:

15% (low)

Question Stats:

72% (01:10) correct 28% (01:23) wrong based on 95 sessions

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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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MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
"Only $99 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" PS Forum Moderator Joined: 25 Feb 2013 Posts: 1216 Location: India GPA: 3.82 What of the followings is the range of X = 1 + (1/2^2) + (1/3^2) [#permalink] ### Show Tags 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 Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6390 GMAT 1: 760 Q51 V42 GPA: 3.82 What of the followings is the range of X = 1 + (1/2^2) + (1/3^2) [#permalink] ### Show Tags 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 _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$99 for 3 month Online Course"
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Joined: 27 May 2012
Posts: 586
Re: What of the followings is the range of X = 1 + (1/2^2) + (1/3^2)  [#permalink]

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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 $$1 < X < 2$$.

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.
_________________

- Stne

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