If x = 1 + 1/2^2 + 1/3^2 + ... + 1/100^2, which of the following must

Question

If  x = 1 + \frac{1}{2^2} + \frac{1}{3^2} + ... + \frac{1}{100^2} , which of the following must be true?

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

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Rucha.Shukla · Answer

Why expand if it's not correct...is it correct?

If yes, logic I used:

Above consists of proper fractions which when squared / have exponents of higher powers give smaller numbers, greater the denominator, smaller the fraction

Since it's 1+ ... It's more than 1

Hence 1 < x < 2

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Rohan271 · Answer

All fractions after 1/(4^2) become smaller and do not contribute any significant amount to the sum..
Simplified sum is 1+.25+.1+.1(1/4^2 = .0625 rounded off to .1) which is less than 2
So answer is B
1<x<2

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unraveled · Answer

If  x = 1 + \frac{1}{2^2} + \frac{1}{3^2} + ... + \frac{1}{100^2} , which of the following must be true?

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

In this x > 1 always
The fraction part after that would always be less than 1 because the highest value of the fraction is  \frac{1}{2^2} (0.25) - almost sum to 1 but not 1.

Hence, 1 + (~1) < 2
Therefore, 
1 < x < 2

Answer B.

PranavVyas · Answer

X is a sum of positive fractions, starting with 1 and adding smaller fractions for each term
Also less than 2, because the second term s 
1/2^2=1/4 ,which is less than 1 Continuing this pattern, x is less than 3, 4, and 5.

IMO, 
B. 1 < x < 2

VivekSri · Answer

OA is B

ruis · Answer

Is there another approach to this question? 
Are there properties for the sum of reciprocals with even denominator?

Bunuel · Answer

Using logic and approximation is the best way to solve this problem.

Tungnguyen218 · Answer

- Since:
x = 1 + 1/2^2 + 1/3^2 + ... + 1/100^2
x < 1 + 1/(1*2) + 1/(2*3) + ... + 1/(99*100)
x < 1 + (1 - 1/2) + (1/2 - 1/3) + ... + (1/99 - 1/100) => we cancel out 1/2 .... 1/99
x < 2 - 1/100
x < 2

- x > 1, obvious but similar to approach above, we can conclude that x > 3/2
x > 1 + 1/2 - 1/101
x > 3/2

=> Answer is 1 < x < 2

WizakoVishwajit · Answer

An average GMAT aspirant (who is not from math background) would mostly know AP and GP to solve Sequences and Series. Assuming that, we can try to correlate given Series with what we know. Like a Series in Geometric Progression 1 + 1/2 + 1/4 + 1/8 + 1/16 ... Where First Term = 1 and Common Ratio = 1/2

Sum of infinite terms of above GP = (First Term)/(1- Common Ratio) = 1 / (1 - 1/2) = 2
If we compare terms of the above series with the terms of given series the given series would be smaller than this GP we can take as a reference (up to 6 terms and then the values will be too small to matter). Hence, we can conclude that the given series is less that 2 and as something positive is added to 1 it has to be more than 1.

Hence, it is 1 < x <2

Or the other approach would be as given in the picture

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If x = 1 + 1/2^2 + 1/3^2 + ... + 1/100^2, which of the following must

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GMAT Problem Solving (PS) Questions

If x = 1 + 1/2^2 + 1/3^2 + ... + 1/100^2, which of the following must

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

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Perfect 805 Score Debrief by Julia

My Rewards