Question

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

Answer 1

Answer: 98

Step-by-step explanation:

Let f(x) = $\frac{x^{2} }{x^{2} -1}$

Find the largest integer n so that f(2) * f(3) * f(4) .... * f(n - 1) * f(n) < 1.98

= $\frac{2^{2} }{2^{2}-1 } * \frac{3^{2} }{3^{2}-1 } * \frac{4^{2} }{4^{2}-1 }*... \frac{(n-1)^{2} }{(n-1)^{2}-1 }* \frac{n^{2} }{n^{2}-1 }$

= $\frac{2*2 }{1*3} * \frac{3*3 }{2*4} * \frac{4*4 }{3*5}*... \frac{(n-1)(n-1)}{(n-1)(n) }* \frac{n*n }{(n-1)(n+1)}$

Observe that the right factor of the numerator cancels out the left factor of the denominator in the next term AND the right factor of the denominator cancels out the left factor of the numerator in the next term. You are left with the following:

$\frac{2*n}{n+1} < 1.98$    

2n < 1.98(n + 1)         multiplied (n + 1) on both sides

2n < 1.98n + 1.98      distributed 1.98 into (n + 1)

0.02n < 1.98             subtracted 1.98n from both sides

      n < 99                divided both sides by 0.02

Since all integers must be LESS THAN 99, the greatest integer is 98