Abstract: The sequence problem is a very interesting problem. Many events in life are closely related to this series. This topic focuses on the judgment of arithmetic progression and arithmetic progression, the nature of arithmetic progression, arithmetic progression's proof and mathematical induction commonly used in mathematical proof. Keywords: sum of the first n items of arithmetic arithmetic progression connectives. In the early stage of the development of mathematics, many people have studied the subject of series, especially arithmetic progression. For example, as early as 2700 BC, the Egyptian mathematics "Reint papyrus" recorded related problems. In the late Babylonian clay tablets, there was also the problem of arithmetic progression. One of the questions is that 10 brothers share 100 silver, and the eldest brother has the most, and the same number decreases in turn. Now I know that BaDi got six taels. What is the difference between two adjacent brothers? Series is extracted from life, and many practical problems encountered in daily life, such as loans, interest rates, depreciation, population growth, radioactive material decay, etc., can be described by arithmetic progression and geometric series. However, how to define series in mathematics? Sequence: A series of numbers arranged in a certain order. Representation method: 1 enumeration method: for example, series; 2 analytical method: general term formula and recursive formula to find the general term of series; inductive method, undetermined coefficient method and formula method to observe the classification of series: 1 divided into finite series and infinite series according to the number of terms; 2 divided into bounded series and unbounded series according to the range of values; 3 divided into increasing series according to monotonicity. Decreasing series and constant series (swing series) We often encounter some problems about series in our daily life. Xiao Ming, a fourth-grade classmate, feels that his English performance is very poor. At present, his vocabulary is only five, yes, no, you, me and him. He decided to recite 10 words every day from today, so from today, his vocabulary will increase day by day, in turn: 56544. Xiao Li is a student in the chemistry department of Shihezi University. His English is very good. He passed Grade 4 in his sophomore year. Her current vocabulary is as high as 4500, but later she became addicted to online games. He plans to stop reciting words from today. As a result, he unconsciously forgets 30 words every day. So from today on, her vocabulary is decreasing day by day, which is: 4500, 4470, 4440. ) From the above two cases, we get two series 15, 15, 25, 35, … and 24500, 4470, 4440, 44 10, … Let's take a closer look at what the above two series have in common * * *: from the second. (Error: the difference between every two adjacent items is equal-it should be specified that the order of the difference is the last item minus the previous item), so we named the series with this feature-arithmetic progression) 1. Arithmetic progression's definition: If a series, starting from the second term, the difference between each term and its previous term is equal to the same constant, we call this series arithmetic progression 2. Arithmetic progression's general term formula: or arithmetic progression is defined by the relationship between two adjacent terms of a series. If the first term of arithmetic progression is and the tolerance is D, according to its definition, we can get that the general term of arithmetic progression can be summarized as follows: Arithmetic average term: If three numbers are arithmetic progression, the middle term is called arithmetic average term ∴ If a series is known as arithmetic progression, we only need to know its first term and tolerance D, and then we can get its general term. Let's specifically study some problems in arithmetic progression. A, arithmetic progression's judgment method 1. If the difference between each item in the series and the previous item from the second item is the same constant d, that is, -=d (constant) is arithmetic progression, and its tolerance is D2. If the series starts from the second term, twice of each term is equal to the sum of the previous term and the latter term, that is, 2 =+ is arithmetic progression (that is, the arithmetic average term of sum). 3. If the general term of a series is n terms or the first polynomial of a constant, that is = (p, q is a constant), then it is a arithmetic progression, and its first term is a tolerance of 4. If the sum of the first n terms of a series is a quadratic polynomial or the quadratic coefficient of n terms is zero. H is a constant), it is arithmetic progression, and its first term is the tolerance of 5. If the series is a arithmetic progression with a tolerance of d and k is a constant, it is a arithmetic progression with a tolerance of kd. 6. If the series is a arithmetic progression with a tolerance of d, r is a constant and also a arithmetic progression with a tolerance of d, for example, 1. Determine whether the following series is arithmetic progression? If the nth term of tolerance (1) is: (2), the nth term of tolerance (3) and the nth term of tolerance (4) are: solution: (1) because = 5 = it is arithmetic progression, its tolerance is (2) because = it is the number of terms. (3) Because = is a quadratic polynomial with n terms, the coefficient of quadratic polynomial is 3, and the constant term is zero, so it is a arithmetic progression, and its tolerance d=6 (4) is a quadratic polynomial with constant term 1, so it is not the basic formula of arithmetic progression II and some simple solutions of arithmetic progression: (1) = or. As the saying goes, it is good to know three and get two. Example 1. In arithmetic progression, it is known that,,, solution 1:∫, then ∴ solution 2:∫∴ Summary: Example 2 of the second general formula. Enter a arithmetic progression's general term formula in the calculator series, let the S term and T term of the series be sum respectively, and calculate the value. And prove your conclusion solution: the value obtained by calculation is always equal to the tolerance proof: let the first term of arithmetic progression be, the last term be, and the tolerance be D, and (1)-(2) get a summary: ① this is the deformation of the second general term formula, ② the geometric characteristics, the slope of the straight line, and ② the solution of the continued term is called the arithmetic average term. If three numbers are passed, it becomes arithmetic progression. If three numbers become arithmetic progression, we usually set the arithmetic mean term as A and the tolerance as D, then these three numbers are:, so their sum is a number whose difference has nothing to do with the public D (only related to the arithmetic mean term A), which can usually simplify the operation. Similarly, if four consecutive numbers become arithmetic progression, we usually set it as: example 1; if three numbers become arithmetic progression, the sum of the three items is 27. Because three numbers become arithmetic progression, the solution can be set to, assuming that the arithmetic progression obtained by knowledge is: 3 9 15 or 15 9 3 cases. Three four numbers become arithmetic progression, and the sum of squares of four numbers is 94. The product of the first number and the fourth number is less than the product of the second number and the third number 18. Solution: Let four numbers be a-. A+3d stands for ∴, so these four numbers are: 8, 5, 2,-1 or 1, -2, -5, -8 or-1, 2, 5, 8 or -8, -5, -2. Example 1. Let the sum of the first four items of arithmetic progression and the sum of the first 12 items find the minimum value of the sum of the first 20 items of (1) series and the sum of the first n items; (2) Solution of the general term formula of the series: Let the formula for finding the sum of the previous terms be: (1) take a small value. (2) By comparison, we know that arithmetic progression's first term is the general term formula: III. The sum of the properties of arithmetic progression (1) is the same. In arithmetic progression, the sum of two items with the same number is equal, that is, if it is special, the situation is 1. Let arithmetic progression have n items, the sum of the first three items is 24, and the sum of the last three items is 60, all of which are items. Find the first solution of arithmetic progression's term, let arithmetic progression's tolerance be d (using the basic formula), and then get the solution of n=32 (2). The second solution (using the same properties of the sum of terms) can be known from the assumption: because it is: because, therefore, (2) the properties of equidistant groups. If arithmetic progression's tolerance is d, starting from the first term, this series is divided into several arrays with the same number of terms (called equidistant adjacent groups), then the sum of adjacent terms in each group is also a arithmetic progression, and its tolerance is a reminder: (arithmetic progression, the tolerance is a reminder. Solution 1 Because,,,, becomes arithmetic progression, let the tolerance be d, and the sum of the former 10 is:, ∴. ∴ former 1 1. Solution 2 Let arithmetic progression's tolerance be d, then the ∴ sequence becomes arithmetic progression. That's it. ∴ 。 Solution 3 Let the tolerance of arithmetic progression be d, then. And ... by ∴. ∴ The first comment is whether the solution is a series formed by arithmetic progression's uniform piecewise summation or arithmetic progression; The second solution is that the series based on the arithmetic mean of the first n items of arithmetic progression is still arithmetic progression; The third solution is to use the summation definition of sequence and the relationship between two terms in arithmetic progression. Remembering these properties of arithmetic progression can often simplify solving problems. (3) The sum of odd and even terms is arithmetic progression. When the number of items is even 2n, it is: when the number of items is odd 2n- 1, it is: example 1. Let the sum of the first 12 terms in arithmetic progression be the ratio of the sum of even terms to the sum of odd terms. Find the tolerance D solution of this arithmetic progression: the solution 1 (applying the basic formula) is obtained by the hypothesis, and the solution 2 (applying the properties of odd-even sum) is obtained by the hypothesis, so, therefore, d=5 Comments: The first method uses the basic formula of arithmetic progression, which is the easiest to think of, but the calculation process is too complicated, while the second method skillfully applies the properties of odd-even sum in arithmetic progression, and the calculation is simple. (4) The correlation between mathematical induction and arithmetic progression proves that mathematical induction is an "inductive axiom" based on natural numbers. The proof process is as follows: Let m be a subset of the set n of natural numbers, and if ① 1∈M is satisfied, ② when k∈M can be deduced, k+ 1 ∈ m, then m = ② When P(k) is a true proposition, it can be concluded that P (k+ 1) is also a true proposition, so for any natural number n, P(n) is a true proposition. The basic form of mathematical induction: For the proposition P(n) related to all natural numbers, if we can: ① prove that the proposition P( 1) holds. ② Suppose that P(k) is true for any natural number k, and prove that P (k+ 1) is also true. It can be asserted that the proposition P(n) holds true for all natural numbers n. According to natural number set's "minimum number principle" (that is, every nonempty subset of natural number set must have a minimum number), another form of mathematical induction (the second mathematical induction) can be derived: for the proposition P(n) related to all natural numbers, if we can: ① prove that the proposition P( 1) holds true. ② Suppose that for any natural number K, P(n) holds when 1≤n≤k, which proves that P (k+ 1) also holds. Then it can be asserted that the proposition P(n) holds for all natural numbers n, such as 1. The sum of the first n terms of the sequence {an} is s n, and the arithmetic mean of an and 2 is equal to the arithmetic mean of Sn and 2 for all natural numbers n. (1) Write the first three items of the series {an}; (2) Find the general formula of sequence {{an}} (write out the derivation process); Solution: (1) From the meaning of the question, when n= 1, there is (a 1+2)/2= radical sign (2s1) s1= a1?. When A2 > 0, we get a2=6. When n=3, there is (a3+2)/2= radical sign (2 * S3). The solution is a3= 10. So the first three terms of this series are 2,6, 10. (2) Solution: Guess from (1) that the sequence {an} has a general formula an=4n-2. The general term formula of sequence {an} is proved by mathematical induction. In (1), a 1=2 has been found, so the above conclusion holds. ② Assuming n=k, the conclusion holds, that is, ak=4k-2 exists. Judging from the meaning of the question, there is (ak+2)/2= radical sign (2*Sk). Substitute ak=4k-2 into the above formula to get 2k. There are [a(k+ 1)+2]/2= radical sign [2s (k+ 1)], and s (k+ 1) = sk+a (k+ 1) instead of sk = 2k 2. Get [a (k+1)+2] 2/4 = 2 [a (k+1)+2k2] a (k+1) 2-4a (k+1)+4-. Solution: ak+ 1=2+4k. So AK+1= 2+4k = 4 (k+1)-2. That is to say, when n=k+ 1, the above conclusion holds. According to ① and ②, the above conclusion is correct. There was an influenza in a city last year (1 1). The data shows that on the day of 1 1, 20 new people in this city were infected with influenza virus. After that, the average number of new infections per day increased by 50 compared with the previous day. Due to the measures taken by the municipal medical department, the spread of the virus has been controlled. From a certain day, the average number of new infections per day decreased by 30 compared with the previous day, and ended at 165438+ 10/0. In these 30 days, there were 8670 people infected with the virus in this city, and the date was 165438+ 10. And ask about the number of new patients that day. Analysis: If 165438+ 10 has the largest number of new infections on the day of N, it can be seen from the meaning of the question that from 165438+ 10 to the day of N, the number of new infections every day constitutes a arithmetic progression; From the n+ 1 day to the 30th day, every newly infected person constitutes another arithmetic progression. The sum of these two arithmetic progression is the total number of infected people this month. Briefly explain: from the topic, 165438+ 10/to n, the number of new infections every day constitutes an arithmetic progression an, a 1 = 20, D 1 = 50,1650. From the n+ 1 day to the 30th day, the number of newly infected people every day constitutes arithmetic progression bn, b 1=50n-60, d2=-30, BN = (50n-60)+(n-1) (-30) = 20n-30. 165438+1On October 30th, the number of new infections was b30-n=20(30-n)-30=-20n+570. So * * * the number of infected people =8670, which is simplified as n2-6 1n+588=0. In eight villages, each village is only adjacent to two villages, and 256 tons of chemical fertilizer are distributed to them according to the following rules: the income of each village is half of the sum of its two adjacent villages. Trying to prove that there is only one division. Solution: Assume that the chemical fertilizers in eight villages are (tons) respectively. Therefore; ; ……; Arithmetic progression, namely,,,. Therefore, the series is arithmetic progression. Let the tolerance of this series be, because the ninth term is equal to the first term 1, then, so. That is, a constant sequence, therefore. So there is only one way to distribute fertilizer, that is, 32 tons per farm. Example 3. The cans are piled into a hexagonal pile (that is, a regular hexagonal pyramid): the top layer is one, the lower layers are arranged in a regular hexagon, one on each side is gradually increased, and the outer ring of the bottom layer is one on each side to find the total number of cans. Solution: The bottom jar is circled into an arithmetic series, except for the middle jar, with a tolerance of 6. The first item is 6 and the last item is 0, so the total number of cans is (6). At the same time, the related knowledge of geometric series and arithmetic progression produced an important series, which is geometric series. This topic is just a brief introduction. If you want to have a deeper understanding, you can compare it with arithmetic progression and find it yourself. (1), the definition of geometric series If a series starts from the second term and the ratio of each term to its previous term is equal to the same constant, then this series is called geometric series, and this constant is called the common ratio of geometric series. The common ratio is usually represented by the letter q (q≠0). For example, the sequence 5,25,125,625 … is a geometric series, and its common ratio is 5. The definition can also be described as: In the sequence {an}, if, then {an} is geometric progression. Easy to know q≠0. Is the geometric series ①. The writing of this formula may be controversial. If so, students can study whether it is possible. Then ask, can you rewrite it into geometric series? Why not? The formula gives the quantitative relationship between the first term and the first term of a series, but can we determine a geometric series? How many conditions do you need to determine a geometric series? Given the first term and the common ratio, how to find the value of either term? Therefore, it is necessary to study the general formula. (2) General term formula of geometric series: ① Incomplete induction ..., this formula is multiplication, so (7), arithmetic progression's comprehensive application and some simple proof examples of general series 1, find the sum of the first few terms of the series: guess: let the sum of the first few terms of this series be, through calculation, through existence, general. Proof: Let's leave aside the above conjecture, and now we will prove it by induction. (i) When = 1, it is true; (ii) set to true, that is, then, this means that it is also true. Evidence of (1) and (2) 6. In order to protect a precious cultural relic, the government decided to build a marble retaining wall. When designing, in order to coordinate with the surrounding scenic spots, the amount of marble of the same specification must be calculated according to the following rules: the first floor uses more than half of all marble, the second floor uses more than half of the remaining marble, the third floor … and so on, until the tenth floor just runs out of stone, how many pieces of marble are needed? How many pieces of marble are used for each floor? Solution: If * * * needs marble blocks, then the first floor:; Second floor: third floor: …; The tenth floor: therefore, there is. Answer: * * * 2046 pieces of marble are used, and each floor uses 1024, 5 12, 256, 128, 64, 32, 16, 8, 4 and 2 pieces respectively. (8) The Prince of Mathematics learned from the Prince of Mathematics Gauss's story "Teacher, I am not fooling around"-"The Prince of Mathematics" Gauss's story When he was seven years old, Little Gauss went to elementary school. The teacher's name is Butner, and he is a famous local mathematician. This young teacher from the city always thinks that children in the countryside are idiots and their talents can't be displayed. In a math class in grade three, Butner lost his temper with the children again, and then wrote a long list of formulas on the blackboard: 81297+81495+81693+…+100701+69. "wow! How many figures are there in total? How to calculate? " The more nervous the students are, the more they can't figure out how to calculate. Butner is very proud. He knew that the last number was 100, which was larger than the previous number 198. Even if these naughty students do their calculations obediently all morning, they won't work out the results. Unexpectedly, in a short time, Gao Xiaosi came over with a small slate with the answer written on it and said, "Teacher, I have finished the calculation." Without looking up, Butner said angrily, "Go, don't be ridiculous. Anyone who wants to scribble numbers must be careful! " Say that finish, waved a hammer fist. But little gauss insisted on not leaving and said, "teacher, I'm not kidding." And gently put the small slate on the platform. Butner took one look and was too surprised to speak. Unexpectedly, this 10-year-old child worked out the correct answer so quickly. It turned out that little Gauss did not add one by one like other children, but observed carefully, used his head and found the law. He found that the sum of two numbers at the beginning and the end of a figure is the sum of 182 196,50, and182196 can be quickly calculated by multiplication. From our present point of view, in fact, the series given by the teacher is the key research of arithmetic progression. Gauss skillfully used the properties of arithmetic progression and got a quick and accurate answer! How interesting the arithmetic series is! (9) Wonderful Fibonacci Series (Beauty is by your side) Fibonacci Leonardo (1, about 1 170- 1250), a medieval Italian mathematician, whose earliest and most important work is The Book of Abacus (6544). The problem is that rabbits are fertile two months after birth. A pair of rabbits give birth to a pair of rabbits every month, and rabbits born two months later also give birth to a pair of rabbits every month. So, starting with a pair of rabbits, how many pairs of rabbits can be cultivated in a year? According to this Lucas, it is not difficult to calculate that the number of rabbits per month constitutes a series of 1, 1, 2, 3, 5, 8, 13, 2 1, 34, 55, 89, 144. The series begins with the third project, each of which is the sum of the first two projects.