Finding patterns in junior high school math and finding patterns in junior high school math problem

The first row has 3 numbers and the last one is 1

The second row has 5 numbers and the last one is 4=1+3=2 squares.

The third row has 7 numbers, and the last one is 9 = 1 + 3 + 5 = 3 squares.

The fourth row has 9 numbers, and the last one is 16 = 1 + 3 + 5 + 7 = 4 squares.

The nth row has 2n+1 numbers, and the last one is 1+3+5+(2n+1)=n squared.

45> 2014 Open Squared》44 2014 in line 44, line 44 has a total of 89 numbers, because 44 square = 1936 45 square = 2025, 2014-1936 = 78

2014 is the 79th number from the right in line 44, and the 11th from the left because there are 89 numbers in line 44

In summary, 2014 appears in line 44, the 11th number from the left.

Look at the right part of the equation.

Row k to the right is exactly the square of k.

The square of 44 1936 is less than 2014, and the square of 45 of 2025 is greater than 2014, so 2014 is on line 44.

And the first digit of the row = 2025-1 = 2024, ranking down the 11th digit of 2014 in the row.

A44 11

The 11th number from the left in line 44.

The first row has 3 numbers, the second row has 5, and the third row has 7 ,......And so on, the nth row has 2n+1 numbers.

Then the first n rows have 3+5+7+....+2n+1)=n(n+2) number. When the calculation shows that there are 1935 numbers in the first 43 lines and 2024 numbers in the first 44 lines, 2014 is in line 44.

The first number in line 44 is 2024 and the second number is 2023, so the 11th number is 2014.

The result of the nth row is the square of n, the first number from the left is n(n+2), the row that appears in 2014 has n 2<=2014<=n(n+2), so n=44, and the first number is 44 46=2024, so 2014 can be known as the 11th number in descending order.

Answer: The number 2014 appears on line 44, the 11th number from the left.

The elite of primary and secondary schools is leveraged.

Math Problem Solving Skills for Finding Patterns in Junior High School:

Find the essence of the problem: find the correspondence between the numbers in the sequence and their serial numbers.

1. Equal difference type.

Compare each number with its predecessor, and if the difference is constant, a constant (often called tolerance), then the nth number can be expressed as an=a1+(n-1)d, where a1 is the first number in the series, d is the difference, and (n-1)d is the sum of the difference from the first to the nth place.

Example...Find the nth digit.

Untie; From the second number, each number increases by 3 over the previous number, and the difference is 3, so the nth digit is: 3+(n-1) 3=3n.

2. The increase is equal difference.

That is, each increase is compared with the previous increase, and the difference between the increase is always equal and is a constant.

3. Proportional type.

Compare each number with its predecessor, and if the ratio is constant and equal, a constant, then the nth number can be expressed as an=a1qn-1, where a1 is the first number in the series and q is the ratio.

Example...Find the nth digit.

Untie; From the second number, the ratio of each number to the previous number is always 2, so the nth digit is: 3 2n-1.

4. The increase is proportional.

That is, each increase is compared with the previous increase, and the increase ratio is always equal and is a constant.

Example...What is the 8th term in a series?

Solution: From the second bundle onwards, the increment of each number from the previous number is .So the 6th number is 17+24=33, the 7th number is 33+25=55, and the 8th number is 55+26=119.

5. Square type: The number column is the square of each serial number, the square of the serial number + constant, and the square of the serial number - constant.

Example 7: The first few terms of the known sequence are .Find the nth term of the sequence.

Solution: From observation, we can see that the first terms of the series are equal to +1, so it can be deduced that the nth term is n2+1.

Example 8: Observe the following numbers:Try to write the 100th number according to this pattern.

Solution: From observation, the first few terms of the sequence are equal to -1, so it can be deduced that the nth term is n2-1, and the 100th number is: 1002-1 = 9999.

6. Exponents. Example 9: Observe the following numbers...Try to write the 11th number according to this pattern.

Solution: From observation, it can be seen that the first few terms of the sequence are equal to .From this, it can be deduced that the nth term is 2n-1, and the 11th number is: 210 = 1024.

Horizontally, the first number of an even row is exactly the square of the number of rows, and the first number of an odd number of a column is exactly the square of the number of columns.

45 2=2025, 2025 is the first number in column 45, and the first number in the odd number column from top to bottom (how many?). Exactly the number of columns) is sequentially reduced, so 2025 can be reduced by 45 numbers, and 2008 is only 17 fewer than 2025, so he is in the 18th row of 45 columns.

Every 16 numbers make up a block, up to 2000 for a total of 500 rows.

From line 501 onwards, it is as follows.

1st column, 2nd column, 3rd column, 4th column...

Line 501 2001 2002 2009 2010

Line 502 2004 2003 2008 2011

Line 503 2005 2006 2007 2012

Line 504 2016 2015 2014 2013

Line 505 2017 ...

Snake has played with it. The law is like a snake, the odd number column turns counterclockwise, and the even number column turns clockwise.

2008 is 18 rows and 45 columns.

The general term is 1 [n n + (-1) (n + 1)] e.g. n = 1, 1 [1 1 + 1] = 1 2

n=2, 1 [2 2+(-1) (2+1)]=1 3,··

So when n = 7, 1 [7 7+(-1) (7+1)]=1 50

1/50

2 = 1 squared + 1

3 = 2 squared - 1

10 = 3 squared + 1

15 = 4 squared - 1

26 = 5 squared + 1

35 = 6 squared - 1

The seventh should be 7 squared + 1 = 50

The denominators are 1 +1, 2 2-1, 3 3+1, 4 2-1, 5 2+1, 6 2-1, 7 2+1, .,

Thus the seventh term is 1 50 (50 = 7 2 +1).

1 50 The odd denominator is squared + 1 The even denominator is squared -1 The first digit is the square of what is the number You know?

Answer: 2n-1

Squared is denoted by ).

This law is number = n

Bringing it in gives the nth number to increase the n-1st number:

n^-(n-1)^

n^-(n^-2n+1)

2n-1 It's better to ask a few more people, the answer may be a little wrong, if it's wrong, please let me know, I'll figure it out for myself.

The second number = the first number + 3

3rd number = 2nd number + 5

。Nth number = nth-1st number + 2n-1

Put the above equation on top of each other.

Get: nth number = first number + 3 + 5 + 7 + .2n-1, that is, the nth number is more than the first one, s=3+5+7+.2n-1 is a total of n-1 numbers.

Write again s=2n-1+2n-3+2n-5+.3 Both S are the same.

Add 2s = 2n + 2 + 2n + 2 + 2n + 2 + 2 + ...2n+2=(2n+2)(n-1)

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

The law is: all combinations of the nth group, the sum of the horizontal and vertical coordinates, for example: (0,n)(1,n-1).

2,n-2)

..n-2，2)(n-1,1)

n,0) got it, the number of the whole points obtained is n+1

(1) Let n 2=(x+y)(x-y).

x+y=n 2 and x-y=1 can be solved to get y=(n 2-1) 2n 2=x+y=(n 2-1) 2+(n 2-1) 2(2) a 2=(b+c)(b-c) is the law (b-c=1), so a 2=b 2-c 2

c^2=a^2+b^2

3) Combine right triangles to think.

(1) Try the equation about n to show the law you have found that the square of a natural number is equal to n+(n+1).

2) The relationship between a, b, and c: a 2+b = c The sum of the squares of two right-angled sides of the Pythagorean theorem is equal to the square of the hypotenuse.

The answer is 321

Method: 1 squared + 1 = 2

1 squared + 2 = 3

2 squared + 3 = 7

3 squared + 7 = 16

7 squared + 16 = 65

So: 16 squared + 65 = x = 321

Look, hehe.

1 squared + 1 = 2

1 squared + 2 = 3

2 squared + 3 = 7

3 squared + 7 = 16

7 squared + 16 = 65

16 squared + 65 = 321

321 Think 1 squared plus 1 equals 2, 1 squared plus 2 equals 3, 2 squared plus 3 equals 7, 3 squared plus 7 equals 16, 7 squared plus 16 equals 65, 16 squared plus 65 equals 321 (finally ko.)

The number of squares in the nth figure is 1+2+3+4+5+6+...n because the second number is the first number plus 2

The third number is the second number plus 3

The fourth number is the third number plus 4

And so on, 、、、

The nth number is the first number plus the second number until the last 、、、

The first 1 and the second 1+2=3

The third 1+2+3=6

The fourth 1+2+3+4=10

The nth 1+2+3+......n=（1+n）n/2