點(diǎn) 等差數(shù)列、等比數(shù)列的性質(zhì)運(yùn)用（The nature of arithmetic progression and geometric progression）

1、點(diǎn)12 等差數(shù)列、等比數(shù)列的性質(zhì)運(yùn)用（The nature of 12 arithmetic progression and geometric progression）The nature of difficulty of 12 arithmetic progression and geometric progressionThe nature of the arithmetic and geometric series is the concept of arithmetic and geometric series, the general formula, extended befor

2、e the N and formula. Using the characteristics of arithmetic geometric series of problem solving, can often avoid the first and tolerance ratio, so that the problems can be solved in the whole operation, to reach operation is flexible, convenient and quick the purpose of it has always received much

3、attention. In the college entrance examination has been focused on examining the contents of this part.Difficult magnetic field(this assumes) arithmetic sequence an of the first N and 30, before 2M and 100, and its former 3M and _.Case studyA case of 1 known function f (x) = (X2).(1) find the invers

4、e function f (x) f- - 1 (x);(2) a1=1, =f-1 (an) (n, N*), an;(3) let Sn=a12+a22+. +an2, Bn=Sn+1Sn existence of minimal positive integer m, such that for any n, N*, bn have set up? If it exists, find the value of M; if not, explain the reasonThis is the subject of proposition intention: a comprehensiv

5、e and relevant function sequence, this paper focuses on students logical analysis ability, is it assumes topic.Relying on the integration of knowledge: inverse function, sequence recursive formula, sequence arithmetic sequence and basic problems, such as monotony function, knowledge in a furnace, in

6、genious structure, novel form, is a fine comprehensive problem.The error analysis: first asked to examine the inverse function, the domain is the inverse function of the range of the original function, this is an error prone points, (2) to ask for the bridge for an series , is not easy to break.Tech

7、niques and methods: (2) asked by formula =4, constructing arithmetic sequence to obtain an, , i.e. uterus is commonly used techniques for the series -; (3) use the idea of function. QSolution: (1) y=, dreams x0)(2) dreams,L is tolerance of 4 arithmetic progression,A1=1 dreams, dreams =4n3, =+4 (n-1)

8、 an0, R an=.(3) bn=Sn+1 - Sn=an+12=, obtained by bn,Let g (n) = g (n) = dreams in the N in N* is a decreasing function,* g (n) the maximum value is g (1) =5, R m5, the existence of minimal positive integer m=6, the bn 0, 2lg2 - (n - 4) Lg3 = 0, n = =5.5. *Because the N, N*, lgan series visible in th

9、e top 5 and the maximum.- one1. properties of the arithmetic and geometric series is a profound manifestation of the two series of the basic laws, is both quick and convenient tool for solving arithmetic and geometric series of problems, due to the application of consciousness.2., in the application

10、 of nature should pay attention to the premise of nature, and sometimes need appropriate deformation3. the clever use of nature, to reduce the amount of calculation is very important in the calculation of the arithmetic and geometric series,But the basic method and establish a goal consciousness, wh

11、at is needed, ask what, both to make full and reasonable use conditions, but also pay attention to the target, and often can make clever use of the effect of the same nature of problem solving.Wipe out difficult trainingFirst, the multiple-choice question1. (assumes) an geometric series first a1= -

12、1, before the N and Sn, if Sn is equal to ().C.2 D.2Two. Fill in the blanks2. (assumes known) a, B, a+b a, B, arithmetic progression, geometric series and AB, 0logm (AB) 0, S130. a3=12 is known.(1) find the range of tolerance d;(2) point out which of the values in S1, S2, and S12 is the largest and

13、explain the reason6. (* assumes) known sequence of an arithmetic progression, tolerance of D = 0, the part made in the series anA, a,., a,. For a geometric progression, in which b1=1, b2=5, b3=17.(1) finding the general term formula of series bn;(2) remember Tn=Cb1+Cb2+Cb3+. +Cbn, beg7. (assumes) an

14、 bn arithmetic progression, geometric series, a1=b1=1, a2+a4=b3, B2, b4=a3, an and bn were obtained before the N and S10 and T10.8. (* assumes) an arithmetic progression, tolerance of D = 0, an = 0 (n = N*), and akx2+2ak+1x+ak+2=0 (k, N*)(1) proof: when k takes different natural numbers, the equatio

15、n has a common root;(2) if the equation of root were x1, X2,., xn,., confirmation: series of arithmetic progression.Reference answerDifficult magnetic fieldSolution 1: Sm=30, S2m=100 into Sn=na1+d, got:Method two: from knowledge, S3m only needs m a1+, the ma1+ d=70 will be 2 - S3m=210., RMethod: thr

16、ee by arithmetic sequence an before the N and Sn formula, is a quadratic function on N, namely Sn=An2+Bn (A, B is a constant). The Sm=30, S2m=100 to.Hence, S3m=A (3m) 2+B - 3m=210Solution four: S3m=S2m+a2m+1+a2m+2+. +a3m=S2m+ (a1+2md) +. + (am+2md) =S2m+ (a1+. +am) +m = 2md=S2m+Sm+2m2d.By the soluti

17、on, we know d=, and we substitute S3m=210.Method five: according to the arithmetic sequence: Sm, S2mSm, the nature of knowledge has become S3mS2m arithmetic progression, which is: 2 (S2mSm) =Sm+ (S3mS2m)* S3m=3 (=210 S2mSm)Method: Six dreams Sn=na1+d,* =a1+dAcupuncture point (n) is a line on the y=+

18、a1, by three points (m), (2m), (3m) S3m=3 (S2m line to Sm =210.)Method seven: let m=1 S1=30, S2=100, a1=30, a1+a2=100, a1=30 *, a2=70* a3=70+ (=110 7030)* S3=a1+a2+a3=210Answer: 210Training for annihilating difficultiesA 1. analysis: the use of geometric series and properties. According to the meani

19、ng of problems, and therefore, a1= - 1, q = 1,So, according to the geometric sequence properties of known S5, S10, S5, S15, S10,. Has become a geometric series, and its ratio is Q5, q5= or q= - R,.LAnswer: BTwo, 2. analysis: solve the A, B, solve the logarithmic inequality can beAnswer: (- 8)3. anal

20、ysis: the use of S odd /S even = get solutionAnswer: eleventh items, a11=294. solution one: assignment methodSolution two:B=aq, c=aq2, x= (a+b), =a (1+q), y= (b+c), =aq (1+q),=2.Answer: 2Three, 5. (1): according to the meaning of problems with solutions:The solution to the D tolerance range - d a12a

21、13, a1a2a3 0 that. Therefore, in S1, S2, S12,. Sk is the maximum condition: AK = 0 and ak+1 = 0, i.e.So, a3=12 dreams, dreams of d 3 - 2 - RDreams: d 4 * - 3, 5.5 K 7.Because K is a positive integer, so that in k=6, S1, S2,., S 12, S6.Method: two d a2. a12a13, therefore, there is a natural number k

22、if k = 1 12, the AK = 0 and ak+1 0, Sk is S1, S2,., maximum value of S12. By arithmetic sequence nature, when m, N, P, Q and N*,And m+n=p+q, so am+an=ap+aq.: 2a7=a1+a13=S13 0, a7 0, * A6 = A70, the S1, S2,., S12 S6 maximum.Three: according to the solution to:At the minimum, Sn is the largest;Dreams:

23、 d 3, 6 * (5 -) 6.5. and, in a positive integer, when n=6, N - (5 - 2) minimum, so S6 is the largest.Comment: the title of the article (1) asked by establishing inequalities solution is the basic requirements, the difficulty is not high, easy to start. (2) asked a higher degree of difficulty, to the

24、 maximum value of Sk for Sn, k = 1 12, one of the ideas is that Sk necessary and sufficient condition for a maximum value is AK = 0 and ak+1 0, the three way is visible Sn as a quadratic function of N, with matching method can be solved. It examines the equivalent transformation of mathematical thin

25、king and logical thinking ability and computing ability, to better reflect the characteristics of the college entrance examination competency test. And the idea of two is through the arithmetic sequence properties and explore the distribution of the series, watershed, so as to find out the solution.6. solutions: (1) by the a52=a1 A17 (a1+4d), namely 2=a1 (a1+16d) a1d=2d2,Dreams d =