7th grade math problems

First, the number of squares.

1. The number of small squares n1=3n

2. The number of squares in four squares n2=2n-2

n=1,n2=0

n=2,n2=2

n=3,n2=4

n=4,n2=6

3. The number of nine-grid squares n3=n-2, n>=3

n=3,n3=1

n=4,n3=2

n=5,n3=3

Therefore, the total number of squares n=n1+n2+n3=3n+(2n-2)+(n-2)=6n-4

1. n=1, the number of quadrilaterals is 6, which is counted as a(1);

2. n=2, the number of quadrilaterals can be counted like this, 2 times a(1), plus the 5 longest quadrilaterals in the vertical column, and the largest quadrilateral on the periphery is counted as one, that is, a(2)=2*a(1)+5+1;

3. n=3, the number of quadrilaterals can be counted like this, 2 times a(2), plus the 5 longest quadrilaterals in the vertical column, and the largest quadrilateral on the periphery is counted as one, that is, a(3)=2*a(2)+5+1;However, there is a repeat of a(1) in this statistics;

4. n=4, the number of quadrilaterals can be counted like this, 2 times a(3), plus 5 longest quadrilaterals in the vertical column, and the largest quadrilateral on the periphery is counted as one, that is, a(3)=2*a(2)+5+1;But this statistics have a repeat of a(2) once;

n, a(n) = 2 * a (n-1) + 6-a (n-2), where n 3;

a(n)-a(n-1) = a(n-1)-a(n-2)+6, where n 3;

a(n)-a(n-1) = a(n-1)-a(n-2)+6, where n 3;

Let s(n)=a(n+1)-a(n).

s(n) is the first series of equal differences with s(1)=a(2)-a(1)=18-6=12 and tolerance d=6;

s(n)=s(1)+(n-1)d=12+6×(n-1)=6(n+1)=a(n+1)-a(n)

a(n+1)=a(n)+6(n+1)

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

a(3)=a(2)+6×3

a(2)=a(1)+6×2

a(1)=0+6×1

The above format is added on both sides.

a(1)+a(2)+.a(n)+a(n+1)=a(1)+a(2)+…a(n)+6×[1+2+3+..n+1)]

a(n+1)=6×[1+2+3+..n+1)]

a(n)=6×[1+2+3+..n]

Is this a 7th grade question???

When n<3, there are 5n-2 squares.

When n>=3, there are 6n-4 squares.

Because 1 11-1 12 is greater than 0, the absolute value can be removed, =1 11-1 12. 1 11-1 12 + 1 12-1 13-1 14 + 1 13, and 1 11-1 14 = 3 154 after removing the opposite