The first type: series (arithmetic progression and geometric progression)

—— Liu, a special teacher of Beijing No.12 Middle School

Zhang Xiaoying, a special teacher of Tsinghua High School Attached.

Sequence is an important topic in high school mathematics, and it is also a common problem in mathematics competition. The most basic ones in the series are arithmetic progression and geometric progression.

A series is a series of numbers arranged in a certain order. If the functional relationship between the nth term an of series {an} and the number of terms (subscript) n can be expressed by the formula an=f(n), this formula is called the general term formula of this series.

From the function point of view, a sequence can be regarded as a positive integer set N* (or its finite set {1, 2, ... n}), and the general term formula of the sequence is also the analytical formula of the corresponding function.

To understand the competition of series, we must first deeply understand and master the definitions and properties of two basic series, and master the (isomorphic) relationship between them.

I. arithmetic progression

If a series starts from the second term, the difference between each term and its previous term is equal to the same constant. This series is called arithmetic progression, and this constant is called arithmetic progression's tolerance, which is usually represented by the letter D.

The general formula of arithmetic progression {an} is:

an = a 1+(n- 1)d( 1)

The first n terms and formulas are:

(2)

It can be seen from the formula (1) that an is a linear function (d≠0) or a constant function (d=0) of n, and (n, an) is arranged in a straight line. According to formula (2), Sn is a quadratic function (d≠0) or a linear function (d =

In the arithmetic series {an}, the arithmetic mean:

,

The relationship between any two am and an is:

an=am+(n-m)d

It can be regarded as arithmetic progression's generalized general term formula.

From arithmetic progression's definition, general term formula and the first n terms formula, we can also deduce that:

a 1+an = a2+an- 1 = a3+an-2 =…= AK+an-k+ 1，k∈{ 1，2，…，n}

If m, n, p, q∈N*, m+n=p+q, then there is.

am+an=ap+aq

Sm- 1=(2n- 1)an，S2n+ 1 =(2n+ 1)an+ 1

Sk, S2k-Sk, S3k-S2k, …, Snk-S(n- 1)k… or arithmetic progression, and so on.

Second, geometric series

If a series starts from the second term and the ratio of each term to the previous term is equal to the same constant, this series is called geometric series. This constant is called the common ratio of geometric series and is usually represented by the letter Q.

The general formula of geometric series {an} is:

an=a 1 qn- 1

The first n terms and formulas are:

In geometric series, the proportional term:

,

And the relationship between any two terms am and an is an = amqn-m.

If the common ratio q of geometric progression satisfies 0 < ∣ q < ∣ 1, then this series is called infinite recursive geometric progression, which has various kinds.

The formula for the sum of items (also called the sum of all items) is:

From the definition of geometric series, the general term formula and the first n terms formula, we can deduce that:

a 1 an = a2 an- 1 = a3 an-2 =…= AK an-k+ 1，k∈{ 1，2，…，n}

If m, n, p, q∈N*, there are:

ap aq=am an，

Write π n = a 1 a2 ... Ann, and then there is.

π2n- 1=(an)2n- 1，π2n+ 1 =(an+ 1)2n+ 1

In addition, each term is a geometric series with positive numbers, and the same base number is taken to form a arithmetic progression; On the other hand, taking any positive number c as the base and constructing a power energy with an exponent of arithmetic progression, then {energy} is geometric progression. In this sense, we say that a positive geometric series and an arithmetic series are isomorphic.

What matters is not only the definitions, properties and formulas of the two basic sequences; Moreover, the mathematical thinking and wisdom contained in the process of summation are extremely precious, such as "inverted addition" (arithmetic progression) and "dislocation subtraction" (geometric progression).

There are two main problems in series, one is to find the general term formula of series, and the other is to find the sum of the first n terms of series.

Third, examples.

Example 1. Let ap, aq, am, an be P, Q, M, N in the geometric series {an}. If p+q=m+n, prove: apoaq=amoan.

It is proved that if the first term of geometric series {an} is a 1 and the common ratio is q, then

ap=a 1 qp- 1，aq=a 1 qq- 1，am=a 1 qm- 1，an=a 1 qn- 1

So:

ap aq=a 12qp+q-2，am an=a 12 qm+n-2，

Therefore: AP AQ = am+an.

Note: This example is an important property of geometric series and is often used in solving problems. It shows that the product of two terms equidistant from both ends (the first two terms and the last two terms) in geometric series is equal to the product of the first two terms and the last two terms, namely:

a 1+k an-k=a 1 an

The same is true for arithmetic progression: in arithmetic progression {an}, the sum of two terms, such as the distance between two ends, is equal to the sum of the first two terms and the last two terms. Namely:

a 1+k+an-k=a 1+an

Example 2. In arithmetic progression {an}, a4+a6+A8+a10+a12 =120, then 2a9-a 10=

a20 b . 22 c . 24 D28

Solution: a4+a 12=2a8, a6+a 10 =2a8 and known or obtained.

5a8= 120，a8=24

And 2a9-a10 = 2 (a1+8d)-(a1+9d) = a1+7d = A8 = 24.

So choose C.

Example 3. It is known that arithmetic progression {an} satisfies a1+A2+A3+…+A10/= 0, then there is ().

a . a 1+a 10 1 > 0 b . a2+a 100 < 0 c . a3+a99 = 0d . a 5 1 = 5 1

[2000 Beijing Spring College Entrance Examination of Science and Engineering (13)]

Solution: Obviously, a1+A2+A3+…+A10/.

So a 1+a 10 1=0, so a2+a100 = a3+a99 = a1+a1= 0.

Example 4. Let Sn be the first n term of arithmetic progression {an}, S9= 18, An-4 = 30 (n > 9) and Sn=336, then n is ().

A. 16 B.2 1 C.9 D8

Solution: Because S9=9×a5= 18, a5=2, a5+an-4=a 1+an=2+30=32, therefore, n=2 1 choose B.

Example 5. Let arithmetic progression {an} satisfy 3a8=5a 13, and a 1>0 > 0, and Sn is the sum of its first n terms, then the largest of Sn(n∈N*) is (). (1995 National Senior High School LeagueNo. 1)

(A)s 10(B)s 1 1(C)s 20(D)s 2 1

Solution: ∫3 A8 = 5a 13

∴3(a 1+7d)=5(a 1+ 12d)

therefore

Make an ≥ 0 → n ≤ 20; When n > 20, An < 0.

∴S 19=S20 maximum, select (c)

Note: Quadratic function can also be used to find the maximum value.

Example 6. Let the first sum tolerance of arithmetic progression be a non-negative integer, the number of terms is not less than 3, and the sum of terms is 972, then such a series * * * has ().

2 (B)3 (C)4 (D)5。

[1997 The Third Question of the National Senior High School Mathematics League]

Solution: Let the first term of arithmetic progression be A and the tolerance be D, then there is () according to the meaning of the question.

That is, [2a+(n- 1)d]on=2×972 (*)

Because n is a natural number not less than 3 and 97 is a prime number, the value of the number n must be a divisor (factor) of 2×972, which can only be one of 97, 2×97, 972 and 2×972.

If d > 0, then d≥ 1 can be known from formula (*) as 2× 972 ≥ n (n-1) and d ≥ n (n-1), so there can only be n=97, and formula (*) can be changed to: a+48d.

If d=0, the formula (*) becomes: an=972, then (*) also has two sets of solutions.

So there are four arithmetic progression * * * set the conditions for this problem, namely:

49, 50, 5 1, …, 145, (***97 items)

1, 3, 5, …, 193, (**97 items)

97, 97, 97, …, 97, (**97 items)

1, 1, 1, …, 1(***972=9409 items)

So choose (c)

Example 7. Will be odd set {1, 3, 5, ...} according to the (2n- 1) odd group from small to large:

{ 1}, {3,5,7},{9, 1 1, 13, 15, 17},…

(Group 1) (Group 2) (Group 3)

Then 199 1 is in the group.

[199 1 the third question of the national high school mathematics league]

Solution: according to the meaning of the question, the first n groups have odd numbers.

1+3+5+…+(2n- 1)=n2。

And1991= 2× 996-1is the 996th positive odd number.

∵3 12=96 1 0, 0 < x-[x] < 1, then the solution is:

From 0 < x-[x] < 1,

∴[x]= 1，

Therefore, it should be filled in

Example 9. The first term of geometric series {an} a 1= 1536 and the common ratio. If πn is used to represent the product of the first n terms, then the largest πn(n∈N*) is ().

(A)π9(B)π 1 1(C)π 12(D)π 13

[1996 National Senior High School Mathematics League Examination Questions]

Solution: The general term formula {an} of geometric series is the sum of the first n terms.

because

So π 12 is the largest.

Option (c)

Example 10. Let x≠y and two series x, a 1, a2, a3, y and b 1, x, b2, b3, y and b4 be arithmetic progression, then =

[1988 National Senior High School League Examination Questions]

Solution: According to the meaning of the question, y-x = 4 (A2-a1) ∴;

And y-x = 3 (B3-B2) VII.

∴

Example 1 1. Let x, y, Z y and z be real numbers, and 3x, 4y and 5z be geometric progression and arithmetic progression, then the value of is. [1992 National Senior High School Mathematics League Examination Questions]

Solution: Because 3x, 4y and 5z are geometric series, there is.

3x 5z = (4y) 2, that is, 16y2= 15xz ①.

And ∵ into arithmetic progression, so there is (2).

Substitute ② into ① to get:

∵x≠0，y≠0，z≠0

∴64xz= 15(x2+2xz+z2)

∴ 15(x2+z2)=34xz

∴

Example 12. The known sets M={x, xy, lg(xy)} and N={0,∣x∣,y}

And M=N, the value of is equal to.

Solution: From M=N, we know that an element in m should be 0, so that lg(xy) can know xy≠0 meaningfully, thus x≠0 and y≠0, so only lg(xy)=0, xy= 1, m = {x,/kl. If y= 1, then x= 1, m = n = {0, 1, 1} is connected with the anisotropy of elements in the set, so y≠ 1, thus ∣ x ∣. . X= 1 y= 1 (inclusive), x=- 1 y=- 1, M=N={0, 1,-1}

At this moment,

therefore

Note: x, x2, x3, …, X200 1 series; and

Under the condition of x=y=- 1, they are all cyclic sequences with a period of 2, S2n- 1=-2 and S2n=0, so 200 1 is not terrible.

Example 13. It is known that the sequence {an} satisfies 3an+ 1+an=4(n≥ 1) and a 1=9, and the sum of its first n terms is Sn, then inequality () is satisfied.

∣ sn-n-6 ∣ 0, ∴. Therefore, for any 1≤k≤n, there is

Note S = a11+A22+A33+…+ANN ⑤

⑥

⑤-⑤:

that is

Comments: The sum in this question is actually the sum of the first n items of the new series formed by the product of arithmetic progression's an=n and geometric progression's corresponding items. Formula ⑤ Multiplies the common ratio on both sides and subtracts the wrong term, which comes down to the solution of geometric series. This method is the basic method to find the sum of the first n terms of geometric series, which is very useful in solving this kind of problems and should be mastered. Textbook P 137 Review Reference Question 3 B Group 6 is entitled: SUM: S =1+2x+3x2+…+NXN-1; Examination questions of science and engineering in Beijing College Entrance Examination in 2003 (16): It is known that the series {an} is arithmetic progression, A 1 = 2, A 1+A2+A3 = 12. (1) Find the general formula of the series {an}; (II) Let bn = an xn (x ∈ r), and find the first n terms of the sequence {bn} and the formula. All of them run through the application of the method of "subtraction of wrong questions"

The second type: exponential function and logarithmic function-exponential function, logarithmic function, exponential function and logarithmic function in No.12 Middle School of Liubei Normal University are very important contents in high school algebra. It plays an important role in college entrance examination and mathematics competition. It is of great significance to master the concept of exponential logarithm and its operational properties, and to master the properties, images and relations of the inverse functions of exponential function and logarithmic function. I. Exponential concept and logarithmic concept: The concept of exponent is derived from the concept of power. Multiply the same factor by a...a (n) =an to find the power, where n is a positive integer. Starting from junior high school, N is first extended to all integers; Then the power sum root is unified and extended to rational exponent; Finally, the concept of exponent holds in the range of real numbers. Euler pointed out: "Logarithmic source is exponential". Generally speaking, if a (a >; 0, a≠ 1) whose power to b is equal to n, that is, ab=N, then the number b is called the logarithm of n with a base, and it is recorded as: logaN=b, where a is called the base of logarithm and n is called a real number. Ab=N and b=logaN are a pair of equivalent formulas, where A is a given normal number that is not equal to 1. When b is given to n, it is an exponential operation, and when n is given to b, it is a logarithmic operation. The reciprocal operation of exponential operation and logarithmic operation. Second, the properties of exponential operation and logarithmic operation 1. Exponential operation mainly has three properties: ax ay = ax+y, (ax) y = axy, (ab) x = ax bx (a > 0, a≠ 1, b>0, b ≠ 1) 2. There are also three logarithmic operations (properties): (1) loga (Mn) = logam+Logan (2) logam/n = logam-Logan (3) logam = nlogam (n.0, a≠ 1, M>0. Negative numbers and zero have no logarithm; The logarithm of 1 is zero, that is, log a1= 0; = 0; The logarithm of the base is 1, that is, logaa = 1 5. Logarithmic basis exchange formula and its inference: basis exchange formula: logaN=logbN/logba inference 1: logamnn = (n/m) Logan inference 2: 3. Exponential function and logarithmic function y = ax(a >；; 0, and a≠ 1) is called exponential function. Its basic situation is: (1) the domain is all real numbers (-∞, +∞) (2) the range is positive real numbers (0, +∞), so the function has no maximum and minimum, only a lower bound, y >；; 0 (3) is a one-to-one mapping, so there is an inverse function-logarithmic function. (4) Monotonicity is: when a >; Add functions in 1; 00, a ≠ 1), f (x+y) = f (x) f (y), f (x-y) = f (x)/f (y) function y = logax(a & gt；; 0, and a≠ 1) is called logarithmic function. Its basic situation is: (1) the domain is a positive real number (0, +∞) (2) the range is all real numbers (-∞, +∞) (3) the corresponding relationship is a one-to-one mapping, so there is an inverse function-exponential function. (4) Monotonicity is: when a >; 1 is an increasing function, when 00, a≠ 1), f (x y) = f (x)+f (y), f (x/y) = f (x)-f (y), for example 1. Find f (11001)+f (2/1001)+f (3/1001)+…+f. Note that11001+1= 2/100/+999/10065438. And f (x)+f (1-x) = (ax/(ax+√ a))+(a1-x/(a1-x+√ a)) = (ax/(ax+√ a). This inspires us to combine the sum formulas and add them: the original formula = [f (11001)+f (1000/1)]+[f (2) Kloc-0/) Let a=4, which is the fill-in-the-blank question of 1986 senior high school mathematics league: Let f(x)=(4x/(4x+2)), then the formula f (11)+f. (2) If a=9 is selected, then f(x)=(9x/(9x+3)), and the summation formula can be changed to find F (1/n)+F (2/n)+F (3/n)+…+F ((n-) This is 20. Example 2.5log25 is equal to: () (a)1/2 (b) (1/5)10log25 (c)10log45 (d)10log52 solution: 5log. =( 1/5)× 10log25 ∴ Selection (b) Description: Logarithmic identity is used here: A Logan = n (a > 0, a ≠ 1, n > 0) This is Beijing/kloc-. Example 3. Calculation solution 1: First, the real number is deformed by the collocation method of compound quadratic root simplification. Solution 2: Transform the real number by using the basic properties of arithmetic roots, which shows that the proper application of multiplication formula can simplify the complex. Example 4. Compare (122002+1)/(122003+1) with (122003+1)/(1222) Solution: for the size of two positive numbers, quotient is a common method to compare 1, and it is recorded as 122003 = a > 0. Then ((122002+1)/(122003+1)); ( 122003+ 1)/( 12200)。 (( 12a+ 1)/(a+ 1))=((a+ 12)( 12a+ 1))/( 12(a+ 1)2)=(( 12 a2+ 145 a+ 12)/ (12a)= 1, that is, logab=( 1/logba), so lglog3 10 and lglg3 are opposites. Set in part, then g(x) is odd function, and g(t)+g(-t)=0. This idea of holistic treatment skillfully uses odd function's properties to solve problems, and the key lies in carefully observing the structural characteristics of functions and the identity deformation of logarithms.

The third type: quadratic function Quadratic function is one of the simplest nonlinear functions, which has rich connotations. In middle school mathematics textbooks, there are in-depth and repeated discussions and exercises on quadratic functions and quadratic equations, quadratic trinomials and quadratic inequalities and their basic properties. It has a far-reaching influence on modern mathematics, even modern mathematics. It has been the key content of the college entrance examination for many years, and the key questions with it as the core content have also changed year by year. Moreover, the content of quadratic function is also a very important proposition object in national and local high school mathematics competitions. Therefore, we must master the basic properties of quadratic function comprehensively and skillfully. The key to learning quadratic function is to grasp the vertex (-b/2a, (4ac-b2)/4a), and the origin of the vertex embodies the collocation method (y = ax2+bx+c = a (x+b/2a) 2+(4ac-b2)/4a); The translation of the image comes down to the translation of the vertex (y = ax2→ y = a (x-h) 2+k); Symmetry of function (symmetry axis x=-b/2a, f (-b/2a+x)=f (-b/2a-x), x∈R), monotone interval (-∞, -b/2a), [-b/2a,+∞], extreme value ((44) There is the following "block diagram": (unary) quadratic function y=ax2+bx+c (a≠0) → a=0 → (unary) linear function Y = BX+C (B ≠ 0) ↑ (unary) quadratic trinomial AX2+BX+. Unary quadratic equation ax2+bx+c=0(a≠0) → a=0 → Unary linear equation BX+C = 0 (B ≠ 0) ↓↓↓↓ Unary quadratic inequality AX2+BX+C > 0 or ax2+bx+c