If the product of 100 rational numbers is positive, how many negative numbers are there at most? What is the sum of all possible negative numbers There are several numbers, the first number is A1, the second number is A2, the third number is A3, etc., and the nth number is an. If A1 = half, starting from the second number, each number is equal to the reciprocal of the difference between 1 and the number in front of it. (1) Try to calculate: A2 = (A3 = (), A4 = () (2) According to the above calculation results, we can guess: A1998 = (), a2000 = ()

If the product of 100 rational numbers is positive, how many negative numbers are there at most? What is the sum of all possible negative numbers There are several numbers, the first number is A1, the second number is A2, the third number is A3, etc., and the nth number is an. If A1 = half, starting from the second number, each number is equal to the reciprocal of the difference between 1 and the number in front of it. (1) Try to calculate: A2 = (A3 = (), A4 = () (2) According to the above calculation results, we can guess: A1998 = (), a2000 = ()

Math problem 1:
Because negative is positive, if there are even negative numbers, the product is an integer. Therefore, there are at most 100 negative numbers and at least 0 negative numbers, and the number is an arithmetic sequence with difference 2
Then there are (100-0) / 2 + 1 = 51 numbers,
The minimum is 0 and the maximum is 100,
Then the sum is (0 + 100) * 51 / 2 = 2550
Math 2:
Because there are a lot of them, it is estimated that they are looking for regularity
a2=1/(1-1/2)=2
a3=1/(1-2)=-1
a4=1/[1-(-1)]=2
a5=1/(1-2)=-1
a6=1/[1-(-1)]=2
……
And so on, after A2, if n is odd, then an = - 1,
If n is even, then an = 2
So, A2 = 2
a3=-1
a4=2
a1998=2
a2000=2