Simple calculation 2 / 12 * 14 + 2 / 14 * 16 + 2 / 16 * 18 + 2 / 18 * 20 + 2 / 20 * 22 + 1 / 22

Simple calculation 2 / 12 * 14 + 2 / 14 * 16 + 2 / 16 * 18 + 2 / 18 * 20 + 2 / 20 * 22 + 1 / 22

2/12*14+2/14*16+2/16*18+2/18*20+2/20*22+1/22=(1/12-1/14)+(1/14-1/16)+(1/16-1/18)+(1/18-1/20)+(1/20-1/22)+1/22=1/12-1/14+1/14-1/16+1/16-1/18+1/18-1/20+1/20-1/22+1/22=1/12

33 + 2 / 7 + 5 + 4 + 1 / 3-16 + 4 / 5-14 + 1 / 7-10 + 1 / 3

33 2 / 7 + 5 4 / 5 + 4 1 / 3-16 4 / 5-14 1 / 7-10 1 / 3
=(33 2 / 7-14 1 / 7) + (5 4 / 5-16 4 / 5) + (4 1 / 3-10 1 / 3)
=19 1 / 7-11-6
=2 and 1 / 7
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Divide natural numbers 1 to 99 into five groups. If the average of each group is equal, what is the sum of the five averages

Let the number of five groups be A1, A2, A3, A4, A5, and the average is x, then a1 + A2 + a3 + A4 + A5 = 99, X * (a1 + A2 + a3 + A4 + A5) = x * 99 = 1 + 2 + 3 +... + 99 = (1 + 99) * 99 / 2 = 4950, x = 50, then 5 * x = 250

1+2+3+4+…… +98 + 99 =? The sum of natural numbers from 1 to 99

1+2+3+4+…… +98+99
There are three solutions, one is Gauss algorithm, because 1 + 99, 2 + 98, 3 + 97 The sum is 49 * 100 + 50 = 4950
The second is formula method, which is the summation formula of arithmetic sequence: (first term + last term) * number of terms / 2, that is, (1 + 99) * 99 / 2 = 4950
The third is to set 1 + 2 + 3 + 4 + +98 + 99 = s
99+98+97+…… +3+2+1＝s
So 2S = (1 + 2 + 3 + 4 +...) +98+99)+(99+98+97+…… +3+2+1)
=100+100+…… +100＝99*100
So s = 99 * 100 / 2 = 4950
Responder: 023zy - trainee Mage Level 3 8-10 17:30

The 99 natural numbers 1-99 are divided into 10 groups. Given that the average number of each group in the 10 groups is equal, what is the average number

The average is X
The numbers of each group are a, B, C, D, e, F, G, h, I, J, K
Obviously, a + B + C +... + j + k is 99
So ax + BX + CX +. + JX + KX = (a + B + C +... + j + k) x = 99x = 4950
There is
99x=4950
So the average value is 4950 △ 99 = 50

The natural number 1.2.3... 1998.1999 is divided into three groups. If the average of each group is exactly equal, what is the sum of the three averages
If you want a formula, you'll get bonus points first

Sn=1+2+..1999=1999*2000/2=1999000
A/3=B/3=/C/3
A/3+B/3+C/3=Sn/3=1999000/3

The natural numbers 1,2,3,4,5, ··, 98 and 99 are divided into three groups. If the average of each group of books is exactly equal, then the average is ()
Analysis: (1 + 99) △ 2 = 50

But there is a simpler way,
1. They were divided into three groups
2. The average number of books in each group is exactly equal
Since the average of each group is equal and divided into three groups equally, this problem is to find the average of 1,2,3,4,5, ·, 98,99
That is: (1 + 99) △ 2 = 50

The average of the six natural numbers is 7. The average of the first four numbers is 8, and the fourth number is 11. Then the average of the last three numbers is 7______ ．

The sum of the last three numbers is: 11 + (7 × 6-8 × 4) = 21, the average of the last three numbers is: 21 △ 3 = 7, so the answer is: 7

Mathematical problems from a number of continuous natural numbers 1,2,3 The average of the remaining numbers is 19 and 8 out of 9
From some continuous natural numbers 1,2,3 If there are two primes in the three numbers removed, what is the maximum sum of the two primes?

The answer from Baidu: first of all, we should know that the average number of several natural numbers is more than 19, so the total number is about 40, so it is 19 and 8 / 9, so the total number after removing three numbers is a multiple of 9, so it is 36, so it is from 1 to 39, the total is 20 * 39 = 780, and the total number after removing three numbers is 19 and 8 / 9 times 3

If there are 99 continuous natural numbers, and the middle one is a, then what is the average number, sum and first number of these 99 natural numbers,

The average is a, and the sum is 99A. The first number is a-49
99 continuous natural numbers, the middle one is a
The sequence of continuous numbers is. (A-3) (A-2) (A-1) a (a + 1) (a + 2) (a + 3)
A there are 49 numbers in the front and 49 numbers in the back
So the average is a, and the sum is 99A, and the first number is a-49