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# algebra

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Let f(n) = n^2/(n^2 + n - 2).

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

Jan 6, 2021

#1
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Explanation:

f(n) = n^2 / (n-1)(n+2)

So f(2) * f(3) * f(4) * ... = 2^2 / (1*4) * 3^2/(2*5) * 4^2/(3*6) * 5^2/(4*7) and note that starting from the 4^2 term, the 4^2 gets canceled by two of the denominators.

The product largely cancels. To take a deeper look let's look at f(2) * f(3) * f(4) * ... * f(10). We get that 2 * 3 * 10 / (11*12). We generalize that for f(2)*f(3)*...*f(n) the formula is 2*3*(n/(n+1)(n+2)). The 2*3 stays the same from the first two fractions. The second part comes from the terms that don't get canceled at the end.

So we need to find the value of n such that 6n/(n+1)(n+2) (I just simplified terms to get that) is less than 1.98.

6n/(n+1)(n+2) < 1.98  ----  6n < 1.98(n+1)(n+2)  ----  6n < 1.98n^2 + 5.94n + 3.96 ---- 1.98n^2 - 0.06n + 3.96 > 0. But the solution to this is all real numbers! Therefore, there is no largest number n, since f(2)*f(3)*...*f(n) is always less than 1.98.

You can try testing this out. f(2) = 1, and from there the numbers only get smaller. So the total product can't possible exceed 1.

You are very welcome!

:P

Jan 6, 2021