The following table is composed of continuous natural numbers starting from 1, observe the pattern a

Rule: The number of each row is the number of the previous row plus 1, and the number of rows is equal to the line number.

The first number in each row is the last number in the previous row plus 1.

The last number of each row is the sum of the number of all previous rows, including the row.

According to Gauss's formula, the last number in the nth row is 1+2+3+....+n = n×（1+n）/2

Substitute the corresponding line number to get:

1；The second value in the ninth row of the table is: 8 9 2+2=382;Find the sum of all the numbers in line 12:

The sum of the first 12 lines - the sum of the first 11 lines = (6 13) (1+6 13) 2-(6 11) (1+6 11) 2=870

3；Find the first and last digit in line n: n (1+n) 2

The first 8 rows have a total of (1+8)*8 2=36 numbers, the last number in line 8 is 36, and the second number in line 9 is 38

The first 11 rows have a total of (1+11)*11 2=66 numbers, the 12th row is 67 to 78, and is (67+78)*12 2=870

The first n-1 line always has (1+n-1)*(n-1) 2=n*(n-1) 2 numbers, and the first number in the nth row is n*(n-1) 2 +1, and this line has n numbers, so the last one is n*(n-1) 2 +n

Solution ideas: (1) It is not difficult to find that the number of numbers in each line is equal to the number of lines, find the total number of numbers in the first 8 lines, and then write the second number of the 9th line of the first verse;

2) Find the total number of numbers in the first 11 rows, and then write out all the numbers in the 12th row, and add them together to get the solution;

3) Find the total number of all numbers in the first (n-1) row, and then write out according to the number of numbers in each row equal to the number of rows

1) The number of numbers in each row is equal to the number of rows, the first 8 rows have a total of 1+2+3+4+5+6+7+8=36 numbers, the 9th row refers to a number of 37, and the second number is 38;

2) Line 11 has 1+2+3+....+11=

2 = 66 numbers, the sum of all numbers in line 12 is: 67 + 68 + 69 + 70 + 71 + 72 + 73 + 74 + 75 + 76 + 77 + 78 =

3) There are a total number of numbers in the first n-1 row: 1+2+3+....+n-1）=

n(n-1)

2, so, the first digit of the nth line is.

n(n-1)

n2-n+2

2, the last digit is.

n(n-1)

2+n=n(n+1)

Comments: Test points for this question: Regular type: the change of numbers

Test Center Comments: This question is an examination of the law of number change, and it is important to observe that the number of numbers in each row is equal to the number of rows, and it is also important to be proficient in the summation formula

The number of digits in row nth is: 2n+1.

The last digit in the nth row is: n 2 and the first digit is: (n-1) 2+1

The first number in line n is (n-1) 2+1

The last number is n 2

There are 2n-1 numbers in row n.

Let's start with the first column of numbers

is exactly 2+1, then an=n 2+1

How many numbers are in line n?

There are 3 in the first row.

There are 5 in the second row and 7 in the third row.

It's exactly a series of equal differences.

So the nth row has 3+(n-1)*2 (n 1)=2n+1(n 1), so the nth row is n 2+1 as the first term, 1 is the equal difference, and the equal difference series sn=(2n+1)(n 2+1)+1 2(2n+1)(2n+1-1)=(2n+1)(n 2+n+1)(n 1).

The last number in row 8 of the table is 64, which is the square of the natural number 8, and there are 15 numbers in row 8;

is expressed in an algebraic formula containing n; The first number in the nth row is (n-1) +1, the last number is n, and there are 2n-1 numbers in the nth row;

Find the sum of the numbers in row n.

n-1)²+1+(n-1)²+n²=(2n-1)[(n-1)²+1]+(2n-1)(n-1)=n(n-1)(2n-1)+(2n-1)=(2n-1)(n²-n+1)

Solution: The rule is: the first number on the left side of each row is added by 1, 2, and 3,..That is, the first number of each row is 1 2n(n-1)+1(n is a natural number).

So 1, the second value in the ninth row of the table is 1 2x9x(9-1)+1+1=38, 2, the twelfth row has a total of 12 numbers, the first is 1 2x12x(12-1)+1=67, the last one is 67+11=78, the sum of these 12 numbers is (67+78)x12x1 2=864

3. The first number in the nth row is 1 2n(n-1)+1, and the last number is 1 2n(n-1)+n

There are 55 in row 3 of the table

number, the sum of the numbers in row 3 is.

2) The last number in row 8 in the table is:

Shared in line 8.

Number; (3) It is expressed by an algebraic formula containing n: the first number in the nth line is n2-2n+2

n2-2n+2

The last number is.

n2n2 line n.

2n-12n-1 number.

This is Yang Hui Triangle, you can look at the basic content of Yang Hui Triangle, see for yourself, Yiduo is not pressing.

(1) The last number in row 8 in the table is sixty-four, which is the square of the natural number 8, and there are 15 numbers in row 8;

2) Expressed in an algebraic formula containing n: the first number in the nth row is n -2n+2 , the last number is n, and there are 2n - 1 numbers in the nth row;

3) Sum according to the series of equal differences, the sum of the numbers in the nth row.

n²-2n+2 + n²) 2n-1) / 2= (n²-n+1) (2n-1)

2n³ -3n² +3n - 1

（1）64，8，15

2) The square of n - 2n + 2, the square of n, 2n - 1

3) The cubic of 2n - the square of 3n + 3n - 1

You didn't even know the formula for summing the difference sequence in the third year of junior high school, and I learned it in the fifth grade: (first term + last item) term number 2

1 All 1) The last number in row 8 of the table is 64, which is the square of the natural number 8, and there are 15 numbers in row 8.

2) Expressed in an algebraic formula containing n: the first number in the nth row is (n-1) +1, the last number is n, and there are 2n-1 numbers in the nth row;

3) If you take out the middle of each row, you get a new column of numbers 1, 3, 7, 13, 21, 31 ,...Then the difference between the nth and (n-1) numbers is 2 (n-1), where the difference between two adjacent numbers is 24, then these two numbers are in the row of the original number table.

It's all wrong. 25 and 24 in a row.