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Let $$f(x) = \frac{x^2}{x^2 - 1}.$$Find the largest integer $n$ so that $f(2) \cdot f(3) \cdot f(4) \cdots f(n-1) \cdot f(n) < 1.98.$

 

\(Let \;\;\;f(x) = \frac{x^2}{x^2 - 1}.\;\;\;\\\text{Find the largest integer n so that }\\ f(2) \cdot f(3) \cdot f(4) \cdots f(n-1) \cdot f(n) < 1.98.\)

 

 

(I have just written the question properly)

Guest Sep 23, 2017
edited by Melody  Sep 24, 2017
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 #1
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Note that 1.98  can be written as  1 + 98 / 100  = 198/100  = 99/50

 

Also  note that    

 

x^2           =           x * x

______             __________

x^2 - 1             (x - 1) ( x + 1)

 

So we can write

 

2*2        3*3        4* 4                 (n - 1) ( n -1)       n *  n

____ *  _____  * _____ *  ....... *  ___________  * ___________ <  99 / 50

1 * 3      2 * 4      3 * 5                 (n - 2) ( n)           (n - 1) (n + 1) 

 

Note that all the terms in red will be "cancelled" in the process and we will be left with

 

2 n     <    99

____       ___                 multiply both sides by (1/2)   and we have that

(n + 1)      50

 

  n           <          99

______              ___

(n + 1)                100

 

And its obvious that the largest integer is   n  =  98

 

 

 

cool cool cool 

CPhill  Sep 24, 2017

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