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