Deduction: First, let's look at a problem of finding the law of sequence: 1, 3,6, 10, 15.
(an episode: I don't want to break down every number to find a pattern like my classmates, relying on luck and pure intelligence, which is inefficient. Through past experience, I think the value of the nth number must be a quadratic algebraic expression. I am thinking: can the nth number be regarded as a quadratic function of the value of the nth number? Then the general formula of quadratic function is listed: y = ax 2+bx+c, then x = 1, y =1; Substitute X=2, y=3 and x=3, y=6 into the above formula, solve the 3 yuan linear equations to get the values of a, b and c, and then bring them back to the above formula to get the answer. When everyone is really unable to solve this kind of problem, you can use the above ideas to solve the problem, which is also a method. )
Closer to home: We can find that: 3-1= 2,6-3 = 3, 10-6 = 4, 15- 10 = 5.
The difference between the third number and the previous number minus the difference between the previous number and the second number is 1.
This basic arithmetic progression property is our starting point. Below, we only analyze the general formula of this kind of arithmetic progression. The difference between the second digit of this series and 1 digit is different from the difference between each digit from the third digit and the previous digit minus the difference between the previous digit and the previous digit. We call this difference x, and the difference between each number from the third number MINUS the previous number and the first two numbers is D.
List the general form of this series first: serial number: 1.
2
three
four
five
. ordinary
Series: n 1
nitrogen
n3
n4
n5
neural network
Therefore: x=n2-n 1.
d=n3-2n2+n 1
Now let's look at how the numbers after n2 are expressed: First of all, this is easy to understand: N2 = n1+x = n1+N2-n1.
Every number in n3 is equal to the previous number plus x plus the previous number plus 1, 2,3 3...D d.
Therefore: n3 = N2+x+d = N2+N2-n1+d = 2n2-n1+d.
n4 = n3+x+d+d = N2+x+d+x+d+d = N2+2x+3d = N2+2 N2-2n 1+3d = 3n 2-2n 1+3d
n5 = n4+x+d+d+d = N2+2x+3d+x+3d = N2+3x+6d = N2+3n 2-3n 1+6d = 4 N2-3n 1+6d
We find that every formula from n3 has a rule: coefficient of n2: serial number: 3.
four
five
. n
numerical value
2
three
four
n- 1
Coefficient of n 1: serial number: 3
four
five
. n
Value: 1
2
three
n-2
(Don't look at the minus sign for the time being)
D coefficient: serial number: 3
four
five
. n
Value: 1
three
six
( 1/2)n^2-(7/2)n+ 1
Every number from n3 can be represented by a formula containing n2, n 1, d, and we know their coefficients and values, so we substitute them into nN's formula:->
nN =(n- 1)N2-(n-2)n 1+d
Popularization and application: We have just discussed the general formula of arithmetic progression of order 2, so what about the series of other orders? Through the above conclusions, we can directly derive the general formula of the first-order arithmetic progression. Since the first-order arithmetic progression does not contain D, D is equal to 0, and the general formula is nn = (n-1) N2-(n-2) n1. The general formula above the second order must be re-derived, and the idea is the same as my above derivation process, but there are some different solutions. Students who are interested and capable can try it themselves. I guess there is a rule in these general formulas, and we can even find the general formula of arithmetic progression of order n according to these rules. Finally, we can use this general formula to find out all arithmetic progression's laws, no matter what order of arithmetic progression. Perhaps, we can even discover some big secrets in the process of exploring these seemingly unrelated numbers!