In a certain orderarrayIs calledseries, while the sequence {an}The nth term of is expressed by a specific formula (including parameter n), which is calledGeneral formula。It's likefunctionOfAnalytic expressionSimilarly, the corresponding a can be obtained by substituting specific n valuesnThe value of the item.The solution to the general term formula of the sequence is usually obtained through several transformations of its recursive formula[1]。
For a sequence {an}, if the difference between any two adjacent terms is a constant, then the number sequence is an equal difference number sequence, and the difference of a certain value is called a tolerance, which is recorded as d;From the first item aoneTo item n anSum of, recorded as Sn。
Then, the general formula is
Its solution is very important, and the idea of "superposition principle" is used:
formula
Add the n-1 formulas above, and many related terms will be eliminated successivelyequationRemaining a on the leftnOn the right, a remainsoneAnd n-1 d, so we can get the general formula above.
In addition, the sum of the first n terms of the sequence
The specific derivation method is relatively simple. The above similar superposition method can be used, and the iterative method can also be used. I will not repeat it here.
It is worth noting that,
, that is, the sum S of the first n itemsnDivide by n to get aoneIt is a new sequence with d/2 as the first term and tolerancenThe problem of sequence of numbers can be easily solved.
Proportional sequence
For a sequence {an}, if the quotient of any two adjacent terms (that is, the ratio of the two) is a constant, then the number sequence is an equal ratio sequence, and this constant value quotient is called a common ratio q;From the first item aoneTo item n anSum of, recorded as Tn。
Then, the general formula is
(i.e. aoneMultiply by the (n-1) power of q, which is derived as the idea of "the principle of continuous multiplication":
atwo=aone* q,
athree= atwo* q,
afour= athree* q,
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an=an-1* q,
Multiply the above (n-1) terms, delete the corresponding terms from the left to the right, and the remaining a on the leftn, remaining a on the rightoneAnd the product of (n-1) q, that is, the general term formula[2]。
In addition, when q=1, the sum of the first n terms of the sequence
When q ≠ 1, the sum of the first n terms of the sequence
=
First order sequence
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concept
It is better to include a in the recurrence formula of the sequencenAnd an+1Is called first-order sequence. Obviously, the recurrence formula of the arithmetic sequence is
an=an-1+D, while the recurrence formula of the proportional sequence is an=an-1* q ； These two can be regarded as special cases of first order sequence.Therefore, the first order can be definedrecursive sequence The form is: an+1= A *an+B ···········⊙, where A and B are constant coefficients.So,Arithmetical sequenceIt is a special case of A=1, andProportional sequenceThis is a special case of B=0.
thinking
Basic ideas and methods: compound deformation is a basic sequence (equal difference and equal ratio) model;Stacking elimination;Successive elimination[3]
Idea 1: Original formula composition (proportional form)
Can make an+1- ζ = A * (an-ζ) ········· ① is the deformed form of the original formula ⊙, that is, the value of ζ is calculated by using the method of undetermined coefficient, and a is obtained after sorting out the formula ①n+1= A*an+ζ - A * ζ, this formula can be compared with the original formula,
ζ - A*ζ = B
That is, ζ=B/(1-A)
After replacement, order bn=an-ζ. Then formula ① becomes bn+1=A*bn, that is, it is converted into aone-ζ) As the first item, take A asCommon ratioOfProportional sequence, we can find bnThe general term formula of {an}The general formula of.
Idea 2: Elimination compound (elimination B)
By an+1= A *an+B ···········⊙ Yes
an= A* an-1+B ··········◎
Formula ⊙ minus formula ◎ gives an+1- an= A *（ an- an-1）······③
Order bn= an+1- anAfter that, formula ③ becomes bn= A*bn-1Proportional sequence, b can be foundnThe general formula of, and then get an- an-1=
(of which
Is a function of n), and then use the superposition method to find an
Second order sequence
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concept
By analogy with the concept of first-order recursive sequence, we can define an+2、an+1、anThe recurrence formula of is a second order sequence, and it is obviously more difficult to find the general formula of such sequence than the first order sequence.In order to facilitate the transformation, we can first interpret the simple form of the second order sequence like this[4]：
an+2= A * an+1+B * an, (Similarly, A, B constant coefficients)
thinking
The basic idea is similar to the first order, except that the undetermined coefficient and the corresponding term should be observed when compounding
Original formula compounding: the original formula is deformed to form an+2- ψ * an+1= ω (an+1- ψ * an)
Comparing this formula with the original formula, we can get
ψ+ω=A and - (ψ * ω)=B
The values of ψ and ω can be obtained by solving these two equations,
Order bn= an+1- ψ*an, the original formula becomes bn+1= ω *bnProportional sequence, b can be foundnGeneral formula bn= f (n) ，
That is, an+1- ψ*an=F (n) (where f (n) is a function of n), and this formula just combines the definition of first order sequence, that is, it only contains an+1And anTwo sequence variables, so as to achieve "order reduction", turn "second order" into "first order", and then solve.
Convert the unequal difference sequence and proportional sequence into related equal difference proportional sequence
one
Appropriate operation deformation example: {an}Medium, aone=3 and an+1= antwo, find anSolution: ln an+1= ln antwo= 2 ln an∴{ln an}Is an equal ratio sequence, where the common ratio q=2, and the first term is ln3 ∨ ln an= (2n-1）Ln3 fault
two
Reciprocal transformation method (applicable to an+1= A*an/ (B*an+C), where A, B, C ∈ R)[5]Example: {an}Medium, aone=1，an+1= an/ ( 2an+1) Solution: 1/an+1= ( 2an+1 ) / an= 1/an+2∴{1/an}Is an arithmetic sequence, the first item is 1, and the tolerance is 2 ∨ an= 1 / (2n-1)
three
Method of undetermined coefficient A. The recurrence formula is an+1= p*an+Q (p, q are constants), and the recursive sequence {an+X} is an equal ratio sequence with p as the common ratio, that is, an+1+ x=p*(an+x) Where x=q/(p-1) (or the set formula can be taken apart and equal to the original formula) Example: {an}Medium aone=1，an+1= 3an+4, find anSolution: an+1+ 2 = 3(an+2)∴{an+2} Is the first item of the proportional sequence is 3, and the common ratio is 3 ∨ an= 3n-2B. The recurrence formula is an+1= p*an+ qn(p, q are constants) Conventional deformation, divide both sides by q at the same timen+1Get an+1/ qn+1= (p / q)*( an/qn）+1/q re order bn= an/ qn, we can get bn+1= k*bn+M (where k=p/q, m=1/q), then use the method mentioned in A above to solve C. The recurrence formula is an+2= p*an+1+q*an, (p, q are constants) can make an+2=xtwo, an+1= x , an=1 Solve xoneAnd xtwo, we can get two formulas an+1- x1 *an= x2 *(an- x1 *an-1)an+1- x2 *an=x1*(an- x2*an-1）Then, subtract the two formulas, and the left side can be obtained (k is the coefficient). The right side can be obtained by using the method of proportional sequence, for example: {an}Medium, aone=1, atwo=2, an+2= (2/3)an+1=(1/3) anSolution: xtwo= 2x/3 = 1/3xone=1,xtwo=-1/3 can get equation group an+1- an= - (1/3)* (an- an-1)an+1+(1/3)* an= an+ (1/3)*an-1Solve an=7/4 - 3/4 × (- 1/3) ^ (n-1) D. Recursive formula an+1= p* an+ an+B (a, b, p are constants) can be transformed into an+1+ xn+1+y = p*(an+ xn+Y) Then compare with the original formula, we can get x, y, that is, we can get {an+xn+y} It is an equal ratio sequence with p as the common ratio. For example: {an}Medium, aone=4， an=3an-1+2n-1 (n ≥ 2) solution: original formula=an+ n+1= 3 [an-1+ (n-1)+1]∴{an+N+1} is an equal ratio sequence, q=3, and the first term is 6 ∨ an= 2×3n- n - 1
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The recurrence formula of characteristic root method is an+1= (A*an+B) / (C*an+D) (A, B, C, D are constants) Let an+1= an=X, the original formula is x=(Ax+B)/(Cx+D) (1) If the solution is the same real root xzero, the sequence {1/(an-xzero）}For example: {an}Meet aone= 2，an+1= (2an-1)/(4an+6) , find anSolution: x=(2x-1)/(4x+6)zero= - 1/21/(an+1/2)=1/[(2an-1-1)/(4an-1+6) +1/2]=1/[an-1+ 1/2] +1∴{1/(an+1/2)} is an arithmetic sequence, d=1, the first term is 2/5 ∨ an=5/(5n-3) - 1/2 (2) If the solution yields two different real roots xone,xtwo, construct {(an- xone)/(an- xtwo）}Is the proportional sequence (xone,xtwoThe position of is not in order, and can be changed) Example: {an}Meet aone= 2，an+1= (an+2)/(2an+1) Solution: According to the question, (an-1)/(an+1)=-1/3 [an-1- 1]/[an-1+1] then {(an-1)/(an+1) } is an equal ratio sequence, q=- 1/3, the first term is 1/3 ∨ an= [1 + (-1)n-1(1/3)n] / [1 - (-1)n-1(1/3)n](3) If there is no real root, this sequence may be a periodic sequence, for example: {an}Medium, aone=2, meet an+1= an-1/ an(n ≥ 2) Solution: aone= 2 , atwo= 1/2 , athree= -1 , afour= 2 , afive=1/2...... So an=2 (n MOD 3=1), 1/2 (n MOD3=1), - 1 (n MOD3=0)
Add, subtract
Example: {an}Meet a ₁+2a ₂+3a ∨+......+nan= n(n+1)(n+2)
Solution: Order bn= a₁+ 2a₂+ 3a₃+……+ nan= n(n+1)(n+2)
