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Seventh grade Xueba 123 Math Answers
5. It can be analyzed according to the meaning of the question.

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. )