There are a series of true fractions, which are arranged as follows: 1 / 2, 1 / 3, 2 / 3, 1 / 4, 2 / 4, 3 / 4, 1 / 5, 2 / 5, 3 / 5, 4 / 5 What is the 99th score?

There are a series of true fractions, which are arranged as follows: 1 / 2, 1 / 3, 2 / 3, 1 / 4, 2 / 4, 3 / 4, 1 / 5, 2 / 5, 3 / 5, 4 / 5 What is the 99th score?

﹙1+13﹚×13÷2=91
13+1+1=15
So the 92nd number is 1 / 15, and the 99th number is 8 / 15

There is a series of scores, according to the following rule: 12, 13, 23, 14, 24, 34, 15, 25, 35, 45 (each score is separated by a comma) So the 1999 score is______ ．

According to the observation, we can get the rule: S = 1 + 2 + 3 + +When n = 63, s = 1953; when n = 64, s = 2016, so the denominator of the number 1999 is 64, the numerator is 1999-1953 = 46, and the score is 4664. So the answer is: 4664

There are a series of true scores, arranged as follows: 12, & nbsp; 13, & nbsp; 23, & nbsp; 14, & nbsp; 24, & nbsp; 34, & nbsp; 15, & nbsp; 25, & nbsp; 35, & nbsp; 45 What is the 1001st score?

Looking for the law, we can know that there is one number with 2 as the denominator, two numbers with 3 as the denominator, and 1 + 2 + +44=990，1+2+3+… +45 = 1035, so the denominator of the 1001st number is 46, and 1001 = 990 + 11, so the 1001st fraction is the 11th number from the left of the fraction whose denominator is 46, so its numerator is 11, so 1146 is the truth fraction

One column number 1, 1, 2, 3, 5, 8, 13, 21 Starting from the third term, each term is the sum of the first two terms. What is the remainder of the 2000 term divided by 8?

Because the law of the remainder of this sequence divided by 8 is: 1, 1, 2, 3, 5, 0, 5, 5, 2, 7, 1, 0, 1, 1, 1, 2, 3 Its cycle period is: 1, 1, 2, 3, 5, 0, 5, 5, 2, 7, 1, 0; 2000 △ 12 = 166 8. In the cyclic number, the eighth number corresponds to 5, so the remainder of item 2000 divided by 8 is 5. A: the remainder of item 2000 divided by 8 is 5

1*2/1+1/2*3+1/3*4...2004*2005/1+2005*2006/1+2006*2007/1

Your numerator and denominator are reversed. The fraction line / is equivalent to a division sign, and the denominator is in the lower right corner
simple form
=1-1/2+1/2-1/3+1/3-1/4+…… +1/2006-1/2007
=1-1/2007
=2006 / 2007 (2006 / 2007)

1/2+1/(2*3)+1/(3*4)+.+1/(2004*2005)+1/(2005*2006)

Split term elimination method
According to 1 / N * (n + 1) = 1 / N - 1 / (n + 1)
1/2+1/(2*3)+1/(3*4)+.+1/(2004*2005)+1/(2005*2006)
=1/2 +1/2-1/3+1/3-1/4+...+1/2005-1/2006
=2005/2006

1+2+3+4+.+2004+2005+2006+.+4+3+2+1=

It can be divided into two 1 + 2 + 3 + +Do 2005 and 2006 add up
Then use the formula 1 + 2 + 3 + 4 + +N = n (1 + n) / 2 operation
It's two 2005 * (1 + 2005) / 2 = 2011015 plus 2006
The result is 2 * 2011015 + 2006 = 4024036

1÷2/1÷3/2÷4/3÷5/4÷…… ÷2006/2005

1÷1/2÷2/3÷3/4÷4/5÷…… ÷2005/2006
=1÷(1/2×2/3×3/4×4/5×…… ×2005/2006)
=1÷(1/2006)
=1×2006
=2006

What is the value of (1 / 3 + 1 / 4 + 1 / 5 +,,, + 1 / 2006) + (2 / 3 + 3 / 4 + 4 / 5,,, + 2005 / 2006)?

Using the law of association, we get 2006 * 1-2 = 2004

1/1*2+1/2*3+1/3*4+.+1/2004*2005+1/2005*2006

1/1*2+1/2*3+1/3*4+...1/2004*2005+1/2005*2006
=1-1/2+1/2-1/3+1/3-1/4+.+1/2004-1/2005+1/2005-1/2006
=1-1/2006
=2005/2006