Find the largest $n$ so that $f(2) \cdot f(3) \cdot f(4) \cdots f(n-1) \cdot f(n) < 1.98.$
3 answers
what the answer?
98
We can factor $f(x)$ as
$$f(x) = \frac{x \cdot x}{(x-1)(x+1)}.$$
So if we write out the product, we see that $f(2) \cdot f(3) \cdot f(4) \cdots f(n-1) \cdot f(n)$ is
$$\frac{2 \cdot 2}{1 \cdot 3} \frac{3 \cdot 3}{2 \cdot 4} \frac{4 \cdot 4}{3 \cdot 5} \cdots \frac{(n-1) \cdot (n-1)}{(n-2) \cdot n} \frac{n \cdot n}{(n-1) \cdot (n+1)}.$$
We count that there are two of each factor from 2 to $n,$ inclusive, in the numerator. In the denominator, we count two of each factor from 3 to $n-1,$ inclusive, and one each of 1, 2, $n,$ and $n+1.$ Most of these cancel, leaving
$$\frac{2 \cdot n}{1 \cdot (n+1)} = \frac{2n}{n+1}.$$
We observe that $1.98 = \frac{198}{100} = \frac{2 \cdot 99}{99 + 1}.$ So $n = 99$ is too high. Also, $\frac{2 \cdot 98}{99} < \frac{2 \cdot 99}{100}.$ (We can check this by cross-multiplying to clear the fractions.) So the answer is $n = \boxed{98}.$
$$f(x) = \frac{x \cdot x}{(x-1)(x+1)}.$$
So if we write out the product, we see that $f(2) \cdot f(3) \cdot f(4) \cdots f(n-1) \cdot f(n)$ is
$$\frac{2 \cdot 2}{1 \cdot 3} \frac{3 \cdot 3}{2 \cdot 4} \frac{4 \cdot 4}{3 \cdot 5} \cdots \frac{(n-1) \cdot (n-1)}{(n-2) \cdot n} \frac{n \cdot n}{(n-1) \cdot (n+1)}.$$
We count that there are two of each factor from 2 to $n,$ inclusive, in the numerator. In the denominator, we count two of each factor from 3 to $n-1,$ inclusive, and one each of 1, 2, $n,$ and $n+1.$ Most of these cancel, leaving
$$\frac{2 \cdot n}{1 \cdot (n+1)} = \frac{2n}{n+1}.$$
We observe that $1.98 = \frac{198}{100} = \frac{2 \cdot 99}{99 + 1}.$ So $n = 99$ is too high. Also, $\frac{2 \cdot 98}{99} < \frac{2 \cdot 99}{100}.$ (We can check this by cross-multiplying to clear the fractions.) So the answer is $n = \boxed{98}.$
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