The general term formula of sequence directly expresses the essence of sequence and is an important method to give sequence. The general term formula of series has two functions. First, any term of the series can be found by the general term formula of the series. Secondly, we can judge whether a number is an item of a series and which item it is through the general term formula of a series. Therefore, finding the general formula of series is one of the most common questions in high school mathematics. It not only examines the mathematical ideas of equivalent transformation and reduction, but also reflects students' understanding of series. It has certain skills and is one of the elements to measure students' mathematical quality, so it often permeates the college entrance examination and mathematics competitions. This paper introduces several common methods of finding the general term of series in order to give readers some enlightenment.
First, the general terms of conventional sequences
Example 1: Find the general term formula of the following series.
( 1)2(22— 1),3(32— 1),4(42— 1),5(52— 1),…
(2)- 1×2( 1),2×3( 1),-3×4( 1),4×5( 1),…
(3)3(2), 1,7( 10),9( 17), 1 1(26),…
Solution: (1) An = n (N2- 1) (2) An = n (n+ 1) (- 1) n) (3) An = 2n+1(N2+)
Comments: Carefully observe the structural characteristics of the given data, find out the corresponding relationship between an and n, and write the corresponding expression correctly.
Second, the general term for arithmetic and geometric series
Use the general formula an = a 1+(n- 1) d and an = a 1qn- 1 to write the general term directly, but first find the first term, tolerance and common ratio according to the conditions.
Third, the general terms of swing sequence.
Example 2: Write the sequence 1,-1, 1,-1, …
Solution: an = (- 1) n- 1
Variant 1: Find the sequence 0, 2, 0, 2, 0, 2, …
Analysis and solution: If each item is subtracted from 1, the order is-1, 1,-1, …
So the general formula of series is an =1+(-1) n.
Variant 2: Find the general formula of sequence 3, 0, 3, 0, 3, 0, ….
Analysis and solution: If each item is multiplied by 3(2), the sequence becomes 2, 0, 2, 0, …
So the general formula of series is an = 2 (3) [1+(-1) n-1].
Variant 3: Find the sequence 5, 1, 5, 1, 5, 1, …
Analysis and Solution 1: If each item is subtracted from 1, the series will become 4, 0, 4, 0, …
So the general formula of the sequence is an =1+2× 3 (2) [1+(-1) n-1] =1+3 (4) [1+(-)
Analysis and Solution 2: If you subtract each item from 3, the sequence becomes 2, -2, 2, -2, …
So the general formula of the sequence is an = 3+2 (- 1) n- 1.
Fourthly, the general term of cyclic sequence.
Example 3: Write the sequence 0. 1, 0.0 1, 0.00 1, 0.000 1, …
Solution: an= 10n( 1)
Variant 1: Find the sequence 0.5, 0.05, 0.005, …
Solution: an= 10n(5)
Variant 2: Find the general formula of the sequence 0.9, 0.99, 0.999, ….
Analysis and solution: each term of this series plus each term of the series 0. 1, 0.0 1, 0.00 1, 0.000 1, ... and all terms are 1, so an =/kloc-0.
Variant 3: Find the general formula of the sequence 0.7, 0.77, 0.777, 0.7777, ….
Solution: an = 9 (7) (1-10n (1))
Example 4: Write the sequence 1, 10, 100, 1000, …
Solution: an = 10n- 1.
Variant 1: Find the sequence 9, 99, 999, …
Analysis and solution: Add 1 to each item of this series to get the series 10, 100, 1000, … so an = 10N- 1.
Variant 2: Write the general formula of the sequence 4, 44, 444, 4444….
Solution: An = 9 (4) (10N- 1)
Commentary: In the process of teaching and learning on weekdays, it is necessary to adopt the basic general formula of series, which requires improving the efficiency of classroom teaching and learning, summarizing and reflecting more, paying attention to association and comparative analysis, avoiding analogy, and there is no need to be afraid of the general formula of complex series.
5. Sum the arithmetic and geometric series to find the general term.
Example 5: Find the general formula of the following sequence.
( 1)0.7,0.77,0.777,… (2)3,33,333,3333,…
(3) 12, 12 12, 12 12 12,… (4) 1, 1+2, 1+2+3,…
Solution: (1) an = = 7× = 7× (0.1+0.01+0.001+…+)
= 7×( 10( 1)+ 102( 1)+ 103( 1)+…+ 10n( 1))= = 9(7)( 1- 10n( 1))
(2)an = = 3×= 3×( 1+ 10+ 100+…+ 10n)= 3× 1- 10( 1- 10n)= 3( 1)( 10n- 1)
(3)an = = 12×( 1+ 100+ 10000+…+ 100n- 1 = 100( 1- 100n)= 33(4)(65438+65438
(4)an = 1+2+3+…n = 2(n(n+ 1))
Comments: The key is to clarify the data characteristics of N items according to the changing law of data.
6. Use the accumulation method to find the general term of an = an- 1+f (n).
Example 6: (1) sequence {an} satisfies a 1= 1 and an = an- 1+3n-2 (n ≥ 2), and an is found.
(2) The sequence {an} satisfies a 1= 1, and an = an-1+2n (1) (n ≥ 2) to find an.
Solution: (1) We know from an = an- 1 that an-an- 1 = 3n-2, f (n) = 3n-2 = an-an- 1.
Then an = (an-an-1)+(an-1-an-2)+(an-2-an-3)+... (a2-a1)+a1.
= f(n)+f(n- 1)+f(n-2)+…f(2)+a 1
=(3n-2)+[3(n- 1)-2]+[3(n-2)-2]+…+(3×2-2)+ 1
= 3[n+(n- 1)+(n-2)+…+2]-2(n- 1)+ 1
= 3×2((n+2)(n- 1))-2n+3 = 2(3n 2-n)
(2) We know that an-an- 1+2n (1) is an-an- 1 = 2n (1), and f (n) = 2n (1) = an-an-655.
Then an = (an-an-1)+(an-1-an-2)+(an-2-an-3)+... (a2-a1)+a1.
= f(n)+f(n- 1)+f(n-2)+…f(2)+a 1
= 2n( 1)+2n- 1( 1)+2n-2( 1)+…+22( 1)+ 1 = 2( 1)-2n( 1)
Comments: When f(n)=d(d is a constant), the sequence {an} is arithmetic progression, and the derivation of arithmetic progression's general term formula in the textbook is actually obtained by the method of accumulation.
Seven, use the accumulation method to find the general term of an = f (n) An- 1.
Example 7:( 1) If the known sequence {an} satisfies a 1= 1 and an = n (2 (n-1)) an-1(n ≥ 2), find an.
(2) The sequence {an} satisfies a 1=2( 1) and an = 2n (1) an- 1, so find an.
Solution: (1) is given by the condition an- 1 (an) = n (2 (n- 1)), and f (n) = n (2 (n- 1)).
an = an- 1(an)an-2(an- 1)…a 1(a2)a 1 = f(n)f(n- 1)f(n-2)…f(2)f(2)a 1
= n(2(n- 1))n- 1(2(n-2))n-2(2(n-3))…3(2×2)2(2× 1) 1 = n(2n- 1)
(2)an = an— 1(an)an—2(an— 1)…a 1(a2)a 1 = 2n( 1)2n- 1( 1)…22( 1)2( 1)= 2 1+2+…+n( 1)= 2
Note: If f(n)=q(q is a constant), {an} is a geometric series, and an = f (n) an- 1 series is a generalization of geometric series. In fact, the derivation of the general formula of geometric series in the textbook is derived by the accumulation method.
Eight, use the undetermined coefficient method to find the general term of an = AAN- 1+B sequence.
Example 8: The sequence {an} satisfies a 1= 1, an+ 1+2an = 1, and find its general formula.
Solution: It is known that an+ 1+2AN = 1, that is, an =-2AN- 1+ 1.
Let an+x =-2 (an- 1+x), then an =-2 an- 1-3x, so -3x = 1, so x =-3 (1).
∴ An -3( 1)=-2 (An-1-3( 1))
Therefore, {an-3 (1)} is a geometric series whose common ratio q is -2 and the first term is an-3 (1) = 3 (2).
∴an-3( 1)=3(2)(-2)n- 1=3( 1-(-2)n)
Comments: Generally speaking, if an+x = a (an- 1+x) has an = an- 1+(a- 1) x, then there is.
(a- 1) x = b knows x = a- 1 (b), so an+a-1(b) = a (an-1+a-1(b)).
An =(A 1+A- 1(B))An- 1-A- 1(B)； Particularly, when A=0, {an} is arithmetic progression; When A≠0 and B=0, the sequence {an} is geometric progression.
Generalization: for the general formula an = an- 1+f (n) (A ≠ 0 and A∈R), the general formula can also be solved by the undetermined coefficient method.
Example 9: The sequence {an} satisfies a 1= 1, and an = 2an-1+3n (1) (n ≥ 2), so we find an.
Solution: Let an+X3N (1) = 2 (an+X3N-1(1)), then an = 2an-1+2x3N-1(1).
And from the known an = 2an- 1+3n (1) so 5x= 1, then x=5( 1). So an+5 (1) 3n (1) = 2 (an-1+5 (1) 3n-1(1)).
Therefore, {an+5 (1) 3n (1)} is q=2, and the first term is a1+5 (1) 3 (1) =15 (60
So an+5 (1) 3n (1) =15 (16) × 2n-1,then an =15 (16 )×
Comments: Generally speaking, for the conditions an = AAN- 1+f (n) and an+g (n) = a [an-1+g (n-1)], then there is Ag (n- 1). It is worth noting the correspondence between an+g(n) and an- 1+g (n- 1). Especially, when f(n)=B(B is a constant), it is the aforementioned example 8.
Can this practice be further promoted? For an = f (n) an- 1+g (n) series, can we use the undetermined coefficient method to find the general term formula?
Let's give an example to make an analogy: let an+k (n) = f (n) [an-1+k (n-1)] expand and get it.
An = f (n) an-1+f (n) k (n-1)-k (n), so that f (n) k (n- 1)-k (n) = g (n) theoretically.
Sequence {an} satisfies a 1= 1, and an = 2n (n) an-1+n+1(1). Find its general formula.
In this way, 2n (n) k (n-1)-k (n) = n+1(1) is obtained. Obviously, at present, we can't easily calculate k (n) with the knowledge of high school mathematics.
Nine, find a through Sn.
Example 10: The sequence {an} satisfies an = 5sn-3, so find an.
Solution: let n= 1 and use A 1 = 5an-3. ∴a 1=4(3).As an = 5Sn-3 ......................................................................................................................................................................
Then an- 1 = 5Sn- 1-3. ................
①-② An-An-1 = 5 (sn-sn- 1) ∴ An-An-1 = 5an.
So an =-4 (1) an- 1, then {an} is the geometric series of q =-4 (1) and the first term an=4(3), then an = 4 (3) (-4 (1).
Note: The recurrence relation contains Sn, and the general term formula is usually obtained from the relation between Sn and an (an = sn-sn- 1 (n ≥ 2)). There are two specific ways: one is to transform the relationship between the first n terms and the general terms revealed by the recursive relationship into the relationship between terms, and then get the general terms according to the new recursive relationship. Second, through an = sn-sn- 1, the relationship between the sum of the first n terms and the general term revealed by the recursive relation is transformed into the relationship between the sum of the first n terms and the sum of the first n- 1, and then the general term formula is obtained according to the new recursive relation.
Ten, turn the countdown into arithmetic progression.
Example: 1 1: It is known that the sequence {an} satisfies a 1= 1 and a.
n+ 1=
An+2(2An), find an.
Solution: by a.
n+ 1=
An+2(2an) has a+1(1) = 2an (an+2) = 2 (1)+an (1), which means an+1.
Therefore, the sequence {an( 1)} is a arithmetic progression with the first term a 1( 1)= 1 and the tolerance d=2( 1).
Then an (1) =1+(n-1) 2 (1) = 2 (n+1), so an=n+ 1(2).
Comments: Pay attention to the observation and analysis of the structural characteristics of the topic conditions, modify the given recursive relationship, and make the series related to the sought series (in this case, the series {an( 1)}) an arithmetic or geometric series, and only need to solve the equation to get the general term formula.
XI. Construct the function model into geometric series.
Example 12: it is known that the sequence {an} satisfies a 1=3 and a.
n+ 1=
(an- 1) 2+ 1, find peace.
Solution: according to condition a
n+ 1=
(an- 1) 2+ 1 got an a.
n+ 1- 1=
(an- 1)2
Logarithms on both sides are lg(a
N+1-1) = LG ((an-1) 2) = 2LG (an-1), that is
So the sequence {LG (an- 1)} is a geometric series with the first term LG (a 1- 1) = lg2, and the common ratio is 2.
Therefore, LG (an-1) = lg22n-1= LG.
Then an- 1 = that is, an=+ 1
Comments: By constructing logarithmic function, the degree of order reduction is achieved, and the original recursive relationship is transformed into geometric series.
Twelve, mathematical induction
Example 13: the sequence {an} satisfies a 1=4 and a.
n=4-
An- 1 (4) (n ≥ 2), find an.
Solution: Find the first few items of the series through the recursive relationship, as shown below.
a 1=4=2+ 1(2) a2=4-
a 1(4)=3=2+2(2) a3=4-
a2(4)=3(8)=2+3(2)
a4=4-
a3(4)=2(5)=2+4(2) a5=4-
a4(4)=5( 12)=2+5(2) a6=4-
a5(4)=3(7)=2+6(2)
Guess: The general formula is an=2+n(2). The proof is given by induction.
Obviously, when n= 1, a 1=4=2+ 1(2), the equation holds.
Suppose that when n=k, the equation holds, that is, ak=2+k(2)
Then when n=k+ 1
k+ 1=4-
ak(4)=4-
k(2))k(2)= 4-k+ 1(2k)= 2+2-k+ 1(2k)= 2+k+ 1(2)
According to the inductive principle, an=2+n(2) exists for all n∈N+.
Comments: First, the first few items are calculated according to the recursive relation, and the general formula is summarized by observing the data characteristics, and then proved by mathematical induction.
Thirteen. Comprehensive application
Example 14: series whose items are known to be positive {a
n2=a
N- 12+2 (n ≥ 2), find one.
Solution: by a.
n2=a
N- 12+2 knows one
n2-a
n- 12=2
Then series {a
12= 1 arithmetic progression.
So a
N2 =1+2 (n-1) = 2n-1,which means an=
Example 15: Sequence {a
n+ 1=a
n+6a
N- 1 (n ≥ 2), find one.
Solution: let a
n+ 1+λa
n=μ(a
n+λa
N- 1), and then
n+ 1=(μ-λ)a
n+μλa
n- 1
And one
n+ 1=a
n+6a
N- 1 is the solution or.
When λ=2 and μ=3, a
n+ 1+2a
n=3(a
n+2a
N- 1), i.e.
n+ 1+2a
not applicable
n+2a
n- 1) =3
Then series {a
2+2a
1= 15 geometric series.
So, one
n+2a
N-1=15× 3n-1= 5× 3n is a.
n=-2a
n- 1+5×3n
rise high
n+x 3n =-2(a
N- 1+x 3n- 1) then a.
n=-2a
N- 1-x 3n, so x =- 1.
So, one
n-3n =-2(a
n- 1-3n- 1)
Thus {a
1-3 = 2 geometric series.
So, one
N-3N = 2× (-2) N- 1 is a.
n=3n+2×(-2)n- 1=3n-(-2)n
When λ =-3 and μ =-2, A can also be obtained.
n=3n-(-2)n
Therefore, the sequence {a}
n=3n-(-2)n
Summary: This paper only introduces several commonly used methods to find the general term formula of series. It can be seen that finding the general formula of a series (especially the series given by recursive relation) is indeed very skillful, which has a great relationship with the basic knowledge and skills and basic thinking methods we have learned. Therefore, in the process of teaching and learning on weekdays, we should not only strengthen the study of basic knowledge, basic methods, basic skills and basic ideas, but also pay attention to cultivating and improving mathematics quality and ability. This requires both teachers and students to improve the efficiency of classroom teaching and learning, pay more attention to summary and reflection, association and comparative analysis, avoid analogy, broaden and deepen some seemingly unremarkable basic propositions horizontally and vertically, and reveal their connections and differences with some problems by weakening or strengthening conditions and conclusions, so as to turn them into new propositions. In this way, no matter from the divergence of content or the deepening of problem-solving thinking, we can get the effect of "showing a branch and grafting it into a forest", which is conducive to the formation and development of innovative thinking.