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Item value post-front.

n-1 a(n-1) 6+(n-1)x6=6nn a(n)

The back-front is equal and the tolerance is 6

a2-a1=12

a3-a2=18

a(n)-a(n-1)=6n

The sides are added together.

a2-a1+a3-a2+..a(n)-a(n-1) =12+18+..6n (total n-1 items).

a(n)-a1=12(n-1)+6(n-1)(n-2)/2a(n)=3n²+3n+1

The nth pattern requires 3n +3n + 1 piece.

There is always only one piece in the middle.

The first pattern is: 1+6

The second pattern is: 1+(1+2)*6

The third pattern is: 1+(1+2+3)*6

The sixth pattern is: 1 + (1 + 2 + 3 + 4 + 5 + 6) * 6 = 127 (pieces) then the nth pattern needs to be a chess piece:

1+(1+2+3+……n)*6

1+[n(n+1)/2]*6

1+3n(n+1)

The first Figure 6 1+1=7

The second figure 6 1 + 6 2 + 1 = 19

The third figure 6 1 + 6 2 + 6 3 + 1 = 37

The sixth 6*1+6*2+6*3... 6*6+1, the nth 6*1+6*2+6*3... 6*n+1

Circle area: Pi* (square of radius).

Square: The square of the length of the side.

Rectangle: length * width.

Trapezoidal: (top bottom + bottom bottom) leather stool *height*

Parallelogram: base * height.

Triangle: Bottom *high*

Cone volume: bottom burn suffocation brigade area * height * (1 3).

Sphere volume: (4 3) * pi * (radius of the third power) sphere area: 4 * pi * (radius squared).

Solution: The three groups have the same number of people, 40 in each group, and a total of 120 students in the sixth grade; If the first group of girls is x, then the first group of boys is 40-x; Since the number of girls in the first group is equal to the number of boys in the second group, the number of girls in the second group is 40-x, and the number of boys is x; If the third group of boys is Y, then the girls are 40-Y; From the condition, y (40-x+x+y)=5 13, and y=25, so the number of boys in the sixth grade is 40+x-x+y=65, and the total number of girls is 120-65=55.

There are as many girls in the first group as there are boys in the second group, so the first.

There are 40 boys in the first and second groups, which account for 1-5 13 = 8 13 of the boys in the sixth grade, so the number of boys in the sixth grade = 40 8 13 = 65, and the number of girls = 120-65 = 55

As a correction, the third condition should be that "three fifths of the boys in the third group of boys in the sixth grade".

If the first group of girls is reversed with the second group of boys, the first group is all boys, the second group is all girls, and the three groups are still the same number;

The third group of boys accounts for 3 5 of the total number of boys, then the first group of boys after the transfer accounts for 2 5 of the total number of boys, and also accounts for 1 3 of the total number of students, then the total number of students is regarded as 2 * 3 = 6 shares, with boys accounting for 5 of them and girls accounting for 1 of them.

According to the proportional distribution, there are 40 * 3 6 * 5 = 100 boys and 20 girls.

Severely disproportionate.

Boys in the third group make up 13 out of 5 of the total number of boys in grade 6. This question is not right, how can the boys in the third group be thirteen fifths of the total number of boys in the sixth grade...

3*40=120, it is known that there are as many girls in group 1 as there are boys in group 2, then the number of boys in group 1 is 40 2=20

In this problem, a b represents the multiplication of b consecutive natural numbers starting from a, for example, 2 3 means the multiplication of three consecutive natural numbers starting from 2, which is 2 3 4 (x 3) 2 = 3660

Finally, there is 2, that is, the product of 2 consecutive natural numbers is 3660, and 3660 is decomposed into prime factors, 3660 = 2 2 3 5 61 = 60 61, so x 3 in parentheses is equal to 60

x 3 represents the multiplication of 3 consecutive natural numbers starting from x.

Then decompose 60 into prime factors: 60 = 2 2 3 5 = 3 4 5x is 3

From "car A arrived at place C on the way one hour earlier than car B, and when car B arrived at place C, car A arrived at place B just right", it can be seen that car A took hours to get from C to B.

If the speed of A is 60km h, then the speed of B is 55km Zheng Ji h, and A and B are x km apart.

x/60=(x-30)/55

x=360 km

A, Lao Cong Song B and the two places are 360 kilometers apart.

1. The number of people per unit area of pool A is 30 (15*8) =, and the number of people per unit area of pool B is 200 (25*40) = the comparison of pool A is greater than pool B, and the answer is that pool A is more crowded, choose A

2.Look at the picture, and get that the triangle and the black circle are opposites, exclude a and d, uh, you seem to be missing something in this picture

1.Choose a because a: 15 8 = 120 120 30 people = 4 (each person occupies the place) b: 40 25 = 1000 2 b、

1000 200 = 5 (each person has a place) 5 4 So choose A.

2.b, c can be reversed.

The first question is the number of people per unit area. . .

Question 2: Why is there only five small squares in the picture?

1. A: 30 people 15A*8A=1 person 4A 2B: 200 people 40A*25A=1 person 5A 2 Pool A 1 person accounts for 4A 2, and 1 person in Pool B accounts for 5A 2, so Pool A is crowded, choose A2, choose C, and eliminate the method.

Fig. 1 7 pieces = 1 + 6 1

Fig.2 19 = 1+6 1+6 2

Fig. 3 37 pieces = 1 + 6 1 + 6 2 + 6 3 Figure 4 1 + 6 1 + 6 2 + 6 3 + 6 4 = 61 pieces Fig. 6 1 + 6 1 + 6 2 + 6 3 + 6 4 + 6 5 + 6 6 = 127 pieces.

Figure n 1+6 1+6 2+......6×n=1+6(1+2+3+4+……n)

1+6×(1+n)×n÷2

3n²+3n+1

Solution: It takes 7 chess pieces to put a pattern: 7=1*6+1, and 19 chess pieces to put a second pattern:

19=3*6+1, 37 chess pieces are needed to place the third pattern: 37=6*6+1, and if you put it down in this way, it takes 21*6+1=127 chess pieces to place the sixth pattern, and n(n+1) 2 *6+1 chess pieces are needed to place the nth pattern.

The coefficients of 6 in each item are: 1=0+1,3=1+2,6=3+3,10=6+4,15=10+5,21=15+6.........

Starting with the second term, each term is equal to the sum of its previous term and the number of terms.

an=a(n-1) +n

an-a(n-1)=n

a(n-1)-a(n-2)=n-1

a2-a1=2

Both sides of the equation add up.

an-a1=2+..n

an=a1+2+..n=n(n+1) 2, so the nth pattern requires n(n+1) 2 *6+1 pieces.

Item value post-front.

n-1 a(n-1) 6+(n-1)x6=6nn a(n)

The back-front is equal and the tolerance is 6

a2-a1=12

a3-a2=18

a(n)-a(n-1)=6n

The sides are added together.

a2-a1+a3-a2+..a(n)-a(n-1) =12+18+..6n (total n-1 items).

a(n)-a1=12(n-1)+6(n-1)(n-2)/2a(n)=3n²+3n+1

The nth pattern requires 3n +3n + 1 piece.

The nth pattern requires a chess piece.

a1=7a2=7+12=19

a3=19+18=37

(1) Note that each subsequent picture is surrounded by an extra circle on the previous map, [there are n-1 chess pieces on each of the 6 sides, but be careful not to repeat the calculation].

a4 = a3 + (6 5-6) = 37 + 24 = 61a5 = a4 + (6 6-6) = 61 + 30 = 91a6 = a5 + (6 7-6) = 91 + 36 = 127 (2) General: From each subsequent figure is enclosed on the previous figure by one more circle to obtain the recursive formula:

an=a(n-1)+[6 (n+1)-6] i.e.: an=a(n-1)+6n(n2)Accumulation calculation:

a2-a1=12

a3-a2=18

an-a(n-1)=6n

Obtained by adding the above n-1 formula:

an-a1=12+18+……6n = 3(n+2)(n-1)∴an=3(n+2)(n-1)+7

i.e.: an=3n +3n+1

It's a matter of numbers. 1,7,19,37, the original formula can be converted into 1+(1+2+3+4......+n-1) 6. Remove the one in parentheses.

1+6n*(n-1)/2

1+3n*(n-1)

The required sixth graph, that is, when n=7, then =1+3*7*6=127 requires the nth graph, that is, when n+1, then =1+3n(n+1).

Look at the figure, the minimum number of rows is n+1, the maximum number is 2n+1, the lower half includes the middle line, turn it up, according to, the parallelogram calculates the number, the number of rows is (2n+1+1) 2 The middle line is repeated once, then the total number n=(n+1 +2n+1) (2n+1+1) 2-(2n+1).

3n+2)(n+1)-(2n+1) =3n +3n+1 when n=1 n=7

When n=2, n=19

When n=3, n=37

When n=6, n=127

First of all, there is always only one piece in the middle.

Then the first pattern is: 1+6

The second pattern is: 1+(1+2)*6

The third pattern is: 1+(1+2+3)*6

The sixth pattern is: 1+(1+2+3+4+5+6)*6=127 (pieces), then the nth pattern needs chess pieces: 1+(1+2+3+......n）*6

This is a hexagonal pattern, the outermost number of points in the nth pattern is 6n, notice that each side has n+1 vertices, which adds up to 6(n+1), and then subtract the 6 vertices that are repeated, which is 6n. So there is:

a[n]=a[n-1]+6n

a1 = 7, i.e. a[n] = 3n 2 + 3n + 1

Solution: From the graph, it can be found that the nth graph has (2n+1) rows, and the middle row has (2n+1) pieces. Push outwards from the middle row, reduce each row by one piece, and be symmetrical with respect to the middle row. The layman has (n+1) pieces.

Then the number of pieces in the nth graph sn=(2n+1)+2(2n)+2(2n-1)+.2(n+1)

From the characteristics of the equal difference series, we know that sn=(2n+1+n+1)(n+1) 2+(2n+n+1)n 2=(3n+2)(n+1) 2+(3n+1)n 2

So the sixth diagram requires 127 pieces.

The nth pattern requires (3n+2)(n+1) 2+(3n+1)n 2=(3n 2+3n+1) pieces.

The sixth pattern needs 1+6*(1+2+3+4+5+6)=127

The nth pattern needs 1+6*(1+n)*n 2 to be adopted.

43;(2n-1)*2+(n-1)*2+7;According to the law of the first three figures, the horizontal number, the two ends of the row of the previous figure add one sub, the upper and lower two rows of subs start from three, and the figure is more than one sub, plus the 7 subs in the original diagram, that is.

The total amount of donations is 56 yuan, and the average donation per person is yuan, so the students who donate the most cannot be below yuan, and because the face value of the donation is an integer, we first consider 10 yuan and 2 yuan, that is, the maximum donation is 12 yuan.

Since it is the student who asks for the most donations to donate as little as possible, other students should try to be as close as possible to his donation, according to the question, it can be seen that the closest combination to 12 yuan is 11 yuan, so we consider how many 11 yuan can be matched after removing a 10 yuan and 2 yuan.

The results are as follows:

Max: 10+2=12 yuan.

Second: 10 + 1 = 11 yuan.

Third: 10 + 1 = 11 yuan.

Fourth: 5 + 5 + 1 = 11 yuan.

Fifth: 5 + 2 + 2 + 2 = 11 yuan.

Therefore, the student who donated the most donated at least 12 yuan.

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