1 table, 6 people
Two tables, 10 people.
3 tables, 14 people
Four tables, 18 people ...
Therefore, every additional table will add four people, because 6-4 = 2, 10-2 * 4 = 2. ...
So the number of people (a)=4* the number of tables (b)+2.
(1)(2) The problem has come out.
(1): Bring the numbers of Table 4 and Table 8 (b=4, b=8) into the formula respectively.
a=4*4+2= 18
a=4*8+2=34
(2): a=90 is brought into the original formula.
90=4b+2
4b=88
b=22
6、
1 is the number 1 (2* 1- 1).
3 is the second number (2*2- 1).
5 is the third number (2*3- 1).
The number 4: 5+2=7(2*4- 1)
The number 5: 7+2=9(2*5- 1)
So the nth number is 2n- 1.
If you want to find several numbers, you need to know which number is 20 1 1, and then you can find out the number of numbers in the table. Which row and which column is 201?
Let 20 1 1 be the nth number.
20 1 1=2n+ 1
n= 1005
So 20 1 1 is the number 1005.
And there is also the number 1005 in this table.
A line has eight numbers: 1005/8 = 125...5.
That is to say, after the number 1005 is filled with 125 lines, there are still five lines.
Therefore, the number 1005 (that is, 20 1 1) is in the fifth column, line 126.
7、
Solution: (1)①÷5+2 = 7,
The three digits on the left are 275 and the three digits on the right are 572.
∴52×275=572×25;
The three numbers on the left are 396.
The two digits on the left are 63 and the two digits on the right are 36.
∴63×369=693×36;
So the answer is: ① 275,572; ②63,36;
(2)∵ The ten digits of the two digits on the left are A, and the single digits are B,
The two bits on the left are 10a+b, and the three bits are 10b+ 10 (a+b)+a,
The two bits on the right are 10b+a, and the three bits are 10a+ 10 (a+b)+b,
The general formula is: (10a+b) × [10b+10 (a+b)+a] = [10a+10 (a+b)+
Proof: left = (10a+b) × [10b+10 (a+b)+a]
=( 10a+b)( 100 b+ 10a+ 10 b+a)
=( 10a+b)( 1 10b+ 1 1a)
= 1 1( 10a+b)( 10b+a)，
right =[ 100 a+ 10(a+b)+b]×( 10 b+a)
=( 100 a+ 10a+ 10 b+b)( 10 b+a)
=( 1 10a+ 1 1b)( 10b+a)
= 1 1( 10a+b)( 10b+a)，
Left = right,
So the formula of the general law of "digital symmetry equation" is:
( 10a+b)×[ 100 b+ 10(a+b)+a]=[ 100 a+ 10(a+b)+b]×( 10 b+a)。
8、
① the number1-1(-(-2) (1-kloc-0/))
2 2 (-(-2)^(2- 1))
3 -4 (-(-2)^(3- 1))
observe carefully ...
-(-2) (n-1north)
② the number11/4 (-1/2) (-1+3).
2 - 1/2 (- 1/2)^(-2+3)
3 1 (- 1/2)^(-3+3)
observe carefully ...
North-1/2) (-n+3
③ Number 1-1/4
2 1/2
3 - 1
Observe carefully ... (as opposed to the second group)
North1/2) (-n+3
Eighth number in each line:
1、-(-2)^(8- 1)=-(-2)^7=2^7
2、(- 1/2)^(-8+3)=(- 1/2)^(-5)=-2^5
3、( 1/2)^(-8+3)=( 1/2)^(-5)=2^5
Add three numbers: 2 7+2 5-2 5 = 2 7.
Add up the nth number of each row.
-(-2)^(n- 1)+( 1/2)^(-n+3)+(- 1/2)^(-n+3)
=-(-2)^(n- 1)
Because n is a positive integer
So n- 1 is an integer.
When n- 1 is odd, -(-2) (n- 1) > 0.
When n- 1 is even, -(-2) (n- 1) < 0.
Because-130 < 0
Therefore, when n- 1 is even, the original formula may be equal to 0.
Because -(-2) 8 =- 128
-(-2)^ 10=- 1024
So when the original formula is equal to-130, n- 1 is not an integer.
So the sum cannot be equal to-130.
(digression, according to the question you asked, you don't know much about the law of finding series. I suggest you analyze it carefully, find out the essence, do some questions, or ask the teacher. )