11/2*5+11/5*8+11/11*14+11/14*17

11/2*5+11/5*8+11/11*14+11/14*17

11/2*5+11/5*8+11/11*14+11/14*17=11/3*（3/2*5+3/5*8+3/11*14+3/14*17）=11/3*（1/2-1/5+1/5-1/8+1/11-1/14+1/14-1/17）=11/3*（1/2-1/8+1/14-1/17）=11/3*（3/8 + 3/14*17）=11*(1/8 +1/238)=11*246/(8*238)=1353/...

5. 68, 17, 81, 4, 57, 36, 1 which are composite numbers? Which are odd numbers? Which are both composite numbers and odd numbers?
emergency

Xiaomantou classmate
The total number is 68,81,4,57,36
Odd numbers are: 5,17,81,57,1
Mixed and odd: 81, 57
Good luck!
If you don't understand the suggestion, you can ask again

In 10,11,12,13,14,15,16,17,18,19, what is the ratio of the number of prime numbers to the number of composite numbers and what is the ratio

There are four prime numbers 11, 13, 17 and 19, and the total number is 10, 12, 14, 15, 16 and 18. Their ratio is 4:6, the simplest integer ratio is 2:3, and the ratio is 2 / 3

In the variable sequence, if the number of times of each group is doubled, then the calculated arithmetic mean a remains unchanged, B is doubled, C is doubled, and D cannot be determined

A unchanged
Double the sum and double the number of items

Some problems of sequence
1. If the sequence satisfies an + 1 = 1 / 2 + √ (an-an2) and A1 = 1 / 2, then the sum of the first 2010 terms of the sequence is equal to
2. Given that the sequence satisfies A1 = 0, a (n + 1) = (an - √ 3) / (√ 3, an + 1), then A20=
3. If the sequence is known to satisfy, then the minimum term of the sequence is item -------- and the maximum term is item --------
………… I'm sure I've done it
Under the root of the first question is the square of an an
In the beginning, I made a mistake

I can't understand the root sign in the first question 2. ∵ A1 = 0, ∵ A2 = √ 3, ∵ A3 = 0, so this sequence circulates, so A20 = √ 3, what is three Well, I forgot. Sorry 1. The first is also a cyclic sequence, A1 = 1 / 2, then A2 = 1, then A3 = 1 / 2, and then a cycle, so s2010 = 1 * 1005 + 1 / 2 * 1005 = 4015 / 23

Some problems in sequence
1,1 / an-1 / a (n-1) = 1 / 2 deduces an = Is n = 1 included in an?
2. If the numerator is 1 and the denominator is a fraction greater than 1, no matter how many of them are added, are they all less than 0?
3. Calculate an with n greater than or equal to 2 and substitute it into BN = lgan. Then is B1 obtained when n = 1 or n = 2?

1, excluding;
2, both greater than 0;
Because n is greater than or equal to 2, there is no B1

1. There are four numbers, of which the first three numbers are equal difference series, the last three numbers are equal ratio series, and the sum of the first number and the fourth number is 37, and the sum of the second number and the third number is 36
2. If B is the median of a and C, y is the median of X and Z, and X, y and Z are all positive numbers, prove that (B-C) logmx + (C-A) logmy + (a-b) logmz = 0, where m ＞ 0, and M is not equal to 1
M is the base. I can't get that effect

The first problem: according to the meaning of the problem, we can respectively set the first number as X, the second number as y, then the third number can be expressed as 36-y, and the fourth number as 37-x, so as to get the following binary linear equations: (1) 2Y = x + (36-y) (2) (36-y) ^ 2 = y (37-x) solution has two groups of values, x = 12, y = 16, then the four numbers are: 12

Some problems about sequence of numbers~
(1) In the known arithmetic sequence {an}, A3 + A5 = a7-a3 = 24, then A2 =?
(2) It is known that a, B and C are in equal proportion sequence, where a = 5 + 2 √ 6 and C = 5-2 √ 6, then B =?
(3) It is known that the common ratio of the equal ratio sequence is 2, and the sum of the first four terms is 1, then the sum of the first eight terms is equal to?

（1）、a7-a3=24=4d——》d=6,
a3+a5=24=2a1+6d——》a1=-6,
——》a2=a1+d=0；
（2）、b^2=ac=(5+2√6)(5-2√6)=1
——》b=+-1；
（3）、S4=a1*(1-2^4)/(1-2)=15a1=1——》a1=1/15,
——》S8=a1*(1-2^8)/(1-2)=255a1=255/15=17.

In the bag, there are cards of the same size marked with 1, 2, 3, 4, 5 and 6 respectively. Take any two cards. (1) find out the probability that the product of the numbers of the two cards is even
(2) The difference between the numbers of one or two cards is regarded as the random variable x-ball distribution column and mathematical expectation

1.
There are 6 * 5 = 30 kinds of any two
The product is an even number, that is, there is at least one even number 2, 4, 6 in the two sheets
1 even 1 odd 3 * 3 * 2 = 18 kinds
2 even 3 * 2 = 6 species
Total 18 + 6 = 24 species
So the probability that the product is even is 24 / 30 = 4 / 5

Urgent solution to a series of problems, thank you
It is known that the sequence an is an equal ratio sequence with the first term of 1, and an ＞ 0, BN is an equal difference sequence with the first term of 1, and A5 + B3 = 21, A3 + B5 = 13. Find the first n terms and Sn of the sequence (BN / (2An))

From the meaning of the question, we can list two formulas: Q4 + 1 + 2K = 21, Q2 + 1 + 4K = 13, and get the solution: q = k = 2
From this we can get: BN = 2N-1, an = 2 (n-1) power
The following is simple, using the equal difference ratio equal ratio method to find it