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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
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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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06 Nov 2017, 20:52
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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
[Reveal] Spoiler: OA

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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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07 Nov 2017, 08:18
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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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08 Nov 2017, 23:56
=>

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

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 $$1 < X < 2$$.

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What of the followings is the range of X = 1 + (1/2^2) + (1/3^2)   [#permalink] 08 Nov 2017, 23:56
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