Wang Haiping, Jiangsu
Sequence is an important content of algebra in senior high school, and it is the basis of learning advanced mathematics. It plays an important role in college entrance examination and various mathematical competitions.
I. arithmetic progression and geometric progression
Example 1. A={ increase the common ratio of geometric progression}, B={ decrease the common ratio of geometric progression}, and find a ∩ B.
Solution: Let q∈A be q>0 (otherwise the series is a swing series).
From an+1-an = a1qn-a1= a1qn-1(q-1) > 0, it is obtained that
When a 1 >; 0, then q> 1; When a 1
Example 2. Find the sum of the first n items of the sequence 1, (1+2) and (1+2+22). ...
Analysis: To find the sum of series, it is imperative to find the general term of series and find some laws from it. The general term is the sum of a geometric series. Let the general term of the series be an, then an =1+2+22+...+2n-1= = 2n-1.
sn =(2- 1)+(22- 1)+(23- 1)+…+(2n- 1)
=(2+22+23+…+2n)-n =-n = 2n+ 1-n-2。
Note: Summing with the following commonly used summation formulas is the most basic and important method for summing series.
Sum formula of arithmetic sequence:
2. Sum formula of equal ratio series:
4,
The common summation methods of sequence are: using common summation formulas to sum; Misalignment subtraction summation; Reverse order addition summation; Group summation method; Sum by split term method; Sum by combination method; Use the sum of the general terms of the sequence and so on.
Example 3. The tolerance of arithmetic progression {an} is known as d=, S 100= 145. Let s odd number = A 1+A3+A5+...+A99, S' = A3+A6+A9+...+A99, and find s odd number, s.
Solution: According to the meaning of the question, we can get S odd +S even = 145.
That is, s odd +(S odd +50d)= 145, that is, 2 odd +25= 145, and the solution is s odd = 120.
And from S 100= 145, we get = 145, so we get a 1+a 100=2.9.
S'=a3+a6+a9+……+a99
===== 1.7 33=56. 1.
Description: The whole idea is an effective means to solve the problem of sequence!
Example 4. In the sequence {an}, a 1=b(b≠0), the first n terms and Sn form a geometric series with a common ratio of Q. 。
(1) Verification: the series {an} is not a geometric series;
(2) let bn = a1s1+a2s2+…+ansn, | q |
Solution: (1) Proof: S 1 = A 1 = B is known.
∫{ Sn} into geometric series, common ratio q.
∴sn=bqn- 1,∴sn- 1=b qn-2(n≥2)。
When n≥2, an = sn-sn-1= bqn-1-bqn-2 = b (q-1) qn-2.
Therefore, when q≠ 1, ==q,
And ==q- 1≠q,∴{an} is not a geometric series.
When q = 1 and n ≥ 2, an=0, so {an} is not a geometric series.
To sum up, {an} is not a geometric series.
(2)∵| q | & lt； 1, from (1), it can be known that n≥2, a2, a3, a4, ..., an constitutes a geometric series with a common ratio of q, ∴ a2s2, a3s3, ..., ansn is a geometric series with a common ratio of q2.
∴bn=b2+a2s2( 1+Q2+Q4+…+q2n-4)
∫S2 = bq，a2=S2-S 1=bq-b
∴a2S2=b2q(q- 1)
∴bn=b2+b2q(q- 1)
∵|q|0， 1600[()n- 1]-4000×[ 1-()n]& gt； 0
Simplify，5×()n+2×()n-7 & gt； 0?
Let x = () n, 5x2-7x+2 >; 0? ∴x 1 (get up)? That is, () n4, so keep the minimum value of the above formula n∈N+ at 5.
Therefore, it will take at least five years to make the county's greening rate reach 60%.
Third, induction, conjecture and proof
Example 7. It is known that the sequence {an} satisfies Sn+an=(n2+3n-2) and the sequence {bn} satisfies b 1=a 1.
And bn=an-an- 1- 1(n≥2).
(1) Try to guess the general formula of the sequence {an} and prove your conclusion;
Solution: (1)∫sn+an =(N2+3n-2), S 1 = A 1, ∴ 2a1= (1+3×/kloc.
∴ a1=1-.When n=2, there is +2a2=(22+3×2-2)=4, ∴a2==2-
Guess, the general formula of the sequence {an} is an=n-
(2) If cn=b 1+b2+…+bn, the value of.
When n=3, there are ++3a3=8, ∴a3==3-.
Proved by mathematical induction as follows:
① When n= 1, a 1= 1-=, the equation holds.
② Assuming that n=k and the equation ak=k- holds, then
When n=k+ 1, AK+1= sk+1-sk = [-AK+1]-[-AK],
. ∴2 ak+ 1=k+2+ak，2 ak+ 1=k+2+(k-)，
∴ak+ 1=(k+ 1)- that is, when n=k+ 1, the equation also holds.
To sum up, we know that an=n- holds true for all natural numbers n.
(2)∵b 1=a 1=，bn = an-an- 1- 1 =[n-]-[(n- 1)-]- 1 =。
∴=[ 1-()n]= 1. ∴cn=b 1+b2+…+bn= 1-()n
Example 8. It is known that the sequence {an} satisfies a 1=2. For any n∈N, there is a >: 0, (n+1) an2+an+1-nan+12 = 0. It is also known that the sequence {bn} satisfies: bn=2n- 1+ 1. ..
(i) Find the general term an of the sequence {an} and its first n terms and Sn; ?
(ii) Find the first n terms of the sequence {bn} and t n; ?
(3) Guess the relationship between Sn and Tn, and explain the reasons.
Solution: (n+1) an2+an+1-an+12 = 0. It is a quadratic homogeneous expression about an and an+ 1, so we can get a more obvious relationship between an and an+ 1 by using the root formula, and then we can find out.
(ⅰ)∵an & gt； 0(n∈N), and (n+1) an2+an+1-nan+12 = 0,
∴ (n+ 1)()2+()-n=0。
∴=- 1 or =
∵an & gt； 0(n∈N),∴=.
∴=。
And a 1=2, so an=2n.
∴sn=a 1+a2+a3+……+an=2( 1+2+3……+n)=n2+n.
(ⅱ)∵bn = 2n- 1+ 1，？
∴tn=b 1+b2+b3+……+bn = 20+2 1+22+……+2n- 1+n = 2n+n- 1
(ⅲ)Tn-Sn = 2n-N2- 1。 ?
When n= 1, T 1-S 1 = 0, ∴ t1= s1;
When n=2, T2-S2=- 1,? ∴T2 When n=3, T3-S3=-2. T3 When n=4, T4-S4=- 1,? ∴t4s5；
When n=6, T6-S6=27,? ∴t6>； S6；
Guess: when n≥5, TN >；; Sn。 That is 2n & gtN2+ 1. Prove by mathematical induction:?
1 When n=5, the former has been verified;
2 if n=k(k≥5), the proposition holds, that is, 2k >；; K2+ 1。 Established,
Then when n=k+ 1,
2k+ 1 = 2 2k & gt； 2(k2+ 1)= k2+k2+2≥k2+5k+2 & gt； k2+2k+2=(k+ 1)2+ 1。
That is, the proposition also holds when n=k+ 1(k≥5).
It can be seen from 1 2 above that when n≥5, there is TN >；; Sn。 ;
To sum up, when n= 1, t1= s1; When 2 ≤ n
Note: Note that the growth rate of 2n is faster than n2+ 1. Therefore, in the process of observation and induction, the conclusion that Tn≤Sn cannot be drawn. Just because there is Tn≤Sn from n= 1 to n=4, we should firmly believe that it must exist, which makes 2n >: N2+ 1, thus enabling the observation process to continue.
Example 9. Known function f(x)=x2-3, (x≤-3)
(1) Find the inverse function f-1(x) of f(x);
(2) Remember A 1 = 1, an =-f-1(an-1) (n ≥ 2), please write the values of A2, A3 and A4, and guess the expression of an. Then prove it by mathematical induction.
Solution: (1) Let y=f(x)= x2-3, (x≤-3), y2 = x2-3 (x ≤-), and x =-y2+3.
That is, f- 1(x)= -x2+3 (x≥0).
(2) If a 1= 1 and an =-f-1(an-1) (n ≥ 2 integer), A2 =-f-1(A1) =
a3=3+4=7，a4=3+7= 10。
According to incomplete induction, we can guess: an=3n-2 (n natural numbers).
The following is proved by mathematical induction:
When n= 1, the proposition A1= 3×1-2 =1holds.
Assuming n=k( 1≤k≤n), the proposition holds: ak=3k-2.
Then when n=k+ 1, ak+ 1=-f- 1(ak).
=ak2+3 =3k-2+3 =3k+ 1-2
To sum up, an=3n-2 applies to all natural numbers n 。
Example 10. In the known sequence {an}, a7=4, an+ 1=,.
(I) whether there is a natural number m, so that when n≥m, an
(2) Whether there is a natural number p, so that when n≥p, there is always 13 and one.
Therefore, we can easily get the relationship between an+ 1-2 and an-2.
an+ 1-2=-2 =。
It can be seen that when
One of the methods is to verify one by one, that is, to solve an= through known conditions. From this, we can calculate the 6th to 1 items of this series from a7, and draw a conclusion.
In addition, thanks to the above solutions, we can also consider the following questions: "If an+1>; 2. We can get a copy of>2 "
From an-2=-2=, it is not difficult to know that the above conclusion is correct.
So there is m= 10, so when n≥m, an
(2) The problem is equivalent to whether there is a natural number p, so that when n≥p, there is always an- 1-an+ 1-2 an < 0.
It can be obtained from (i): an- 1-an+ 1-2 an=.
We already know that when n≥ 10, an
Observing the previous calculation results, we can see: a 100, and draw a conclusion.
Explanation: (1) To sum up, conjecture is based on careful observation and analysis, not water without a source, a tree without a root. (2) If the above analysis process is written by mathematical induction, it is quite concise, but it also covers up the thinking process.
Fourthly, the problem of sequence is discussed from the recursive formula.
Example 1 1. Let An be the sum of the first n terms of series {an}, An=(an- 1), and the general term formula of series {bn} is bn=4n+3.
(1) Find the general term formula of sequence {an};
(2) Arrange the general terms of series {an} and {bn} into a new series {dn}, and prove that the general term formula of series {dn} is dn = 32n+1;
(3) Let the nth term of series {dn} be the r term in series {bn}, Br be the sum of the first r terms of series {bn}, dn be the sum of the first n terms of series {dn}, and Tn=Br-Dn.
Solution: (1) From An= (an- 1), we can know that an+1= (an+1-1).
∴ an+1-an = (an+1-an) = an+1,that is =3.
And a1= a1= (a1-1), a 1=3.
So the sequence {an} is a geometric series with the first term of 3 and the common ratio of 3. The general formula of sequence {an} is an=3n.
(2)∵32n+ 1 = 3 32n = 3(4- 1)2n
= 3×(42n+c 12n 42n- 1(- 1)+…+C2n2n- 1 4(- 1)+(- 1)2n)
=4m+3
∴32n+ 1∈{bn}
And the number 32n=(4- 1)2n.
= 42n+c2n 1 42n- 1(- 1)+…+C2n2n- 1 4(- 1)+(- 1)2n
=(4k+ 1)
∴32n{bn}
And the sequence {an} = {32n+ 1} ∨{ 32n}
∴ dn=32n+ 1
(3) From 32n+ 1 = 4 r+3, we know that r=
∫Br = = r(2r+5)= 1
Dn= ( 1-9n)=(9n- 1)
∴Tn=Br-Dn=-(9n- 1)
= between 34n and 32n+
∫(an)4 = 34n
∴=
Example 12. Known function f (x) = x+x2-a2 (a >; 0)
(1) Find the inverse function f-1(x) of f(x) and its domain;
(2) the sequence {an} satisfies a1= 3an+1= f-1an.
Let bn= and the sum of the first n terms of the sequence {bn} is Sn. Try to compare the sum of Sn and prove your conclusion.
Solution: (1) Square both sides of y-x=x2-a2 to get x=
∫y-x = y-= = 0
∴y≥a or -a ≤ y
So f- 1(x)=, and its location is [-a, 0)∩[a,+∞].
(2)∫an+ 1 = f- 1(an)= 1
∴ bn+ 1 = … = () 2 = bn2 (both sides can be solved by logarithm)
And a 1 = 3a, b 1 = =
∴bn=(bn- 1)2=(bn-2)=(bn-3)
=…=(b 1) =()
∴Sn=b 1+b2+…+bn
=+()2+()+[()+()+…+()]= = 1-()n
Therefore, when n
∫2n- 1 =( 1+ 1)n- 1 = 1+c 1n- 1+C2N- 1+C3N-。
Then when n≥4, 2n-1>; 1+c 1n- 1+C2n- 1
= 1+(n- 1)+>n+ 1
∴().
When n=3, Sn =+() 2+() =++=
Therefore, when n≤3, Sn2, so a22, an >；; 2, so {an} decreases monotonously. And because an >: 2, so
An-2 = <; (An-1-2)
& lt()2(an-2-2)& lt； ... …2pq, a 1, b 1 is not zero, ∴ C22 ≠ C 1 C3, so {cn} is not a geometric series.
Note: This title is the 2000 National College Entrance Examination Mathematics Test. There are many proofs, and readers are advised to look at this problem from different angles. We can draw a more general conclusion.
Inference 1: Let the sequence {cn}, cn=an+bn and a≠b, then the necessary and sufficient conditions for the sequence {cn+ 1-pcn} to be a geometric series are p=a or p = b. 。
Inference 2: Let {an} and {bn} be two geometric series, then the necessary and sufficient condition for the sequence {an+bn} to be a geometric series is that the common ratio of the sequence {an} and {bn} is equal.
Inference 3: The common ratio is the geometric series {an} and {bn} of A and B, and A ≠ B, S and T are real numbers that are not all zero. The necessary and sufficient condition for cn=san+tbn to be a geometric series is st=0.
Example 15. In the sequence {an}, a 1=8, a4=2, an+2=2an+ 1-an n∈N is satisfied.
(1) Find the general term formula of sequence {an};
(2) let sn = | a1||| a2 |+…+| an | to find sn;
(3) let BN = (n∈N), TN = B 1+B2+…+BN (n∈N), whether there is a maximum integer m, so that TN >； exists for any n ∈ n; If it exists, find the value of m; If it does not exist, please explain why.
Solution: (1) is an+2=2an+ 1-an.
An+2-an+1= an+1-an, which means that {an} has become arithmetic progression, and d==-2.
-∴an= 10-2n
(2) From an= 10-2n≥0, n≤5.
When n≤5, ∴, Sn=-n2+9n.
When n >; At 5 o'clock, Sn=n2-9n+40.
Therefore, sn =-N2+9n1≤ n ≤ 5n2-9n+40n > 5 (n ∈ n).
(3)bn===()
∴Tn= b 1+b2+…+bn
=[( 1-)+(-)+(-)+……+(-)]=( 1-)=
& gt& gtTN- 1 & gt； TN-2 & gt； ……& gt； T 1。
∴ tn> to be manufactured is always established and needs to be made.