8,9

Find the value of the product (1-1/4)(1-1/9)...(1-1/100) (the numbers in the denominators are equal to the squares of natural numbers from 2 to 10).
Each factor can be brought to a common denominator and expanded using the difference of squares formula.

## Solution

Transform the given expression: $(1-1 / 4)(1-1 / 9) \ldots(1-1 / 100)=\left(\left(2^{2}-1\right) / 2^{2}\right)\left(\left(3^{2}-1\right) / 3^{2}\right) \ldots\left(\left(10^{2}-1\right) / 10^{2}\right)$. Using the difference of squares formula, further transform the last expression to $\left((2-1) *(2+1) / 2^{2}\right)\left((3-1)(3+1) / 3^{2}\right) \ldots$ $\left((10-1)(10+1) / 10^{*} 10\right)=(1 * 3 / 2 * 2)(2 * 4 / 3 * 3) \ldots\left(9 * 11 / 10^{*} 10\right)$. We see that almost all factors in the numerator and denominator cancel out. In the numerator, 11 remains, and in the denominator, 2 and 10 remain. Thus, the value of the given product is $11 / 20=0.55$.

## Answer

0.55.