Type one
Inductive conjecture proof
We can write the first few terms of the sequence from the recursive formula of the sequence, then sum up the rules from the first few terms, guess a general term formula of the sequence, and finally prove it by mathematical induction.
Type 2
"Difference Law" and "Product Commercial Law"
(1) When the recurrence formula of a series can be changed into an+ 1-an=f(n), take n = 1, 2, 3, ..., n- 1, and get n- 1 formulas.
a2-a 1=f( 1)，a3-a2=f(2)，…，an-an- 1=f(n- 1)，
And if f (1)+f (2)+…+f (n-1) can be found, the general term an can be accumulated on both sides. This method is called "step by step method".
(2) When the recurrence formula of a series can be changed into an+ 1/an=f(n), let n= 1, 2,3, …, n- 1, then we can get n- 1 formulas, that is.
A2/A 1 = F ( 1)，A3/A2 = F (2)，A4/A3 = F (3)，…，an/An- 1 = F (n- 1)，F ( 1)
Type 3
structured approach
The recurrence formula is pan=qan- 1+f(n)(p and q are nonzero constants), and a new geometric series solution can be constructed by using the undetermined coefficient method.
Type 4
Can be converted into type three passthrough items.
(1) "Logarithmic method" was transformed into type III.
The recurrence formula is an+ 1=qan? K (q > 0, k ≠ 0 and k ≠ 1, a 1 > 0), take the common logarithm on both sides and get lgan+ 1=klgan+lgq. Let lgan=bn, then we have bn+/kloc-.
(2) Convert "Reciprocal Method" into Type 3.
The recurrence formula is in the form of quotient: an+1= (pan+b)/(qan+c) (an ≠ 0, PQ ≠ 0, PC ≠ QB).
If b=0, an+ 1 = pan/(qan+c). Because an≠0, the reciprocal of both sides is 1/an+ 1=q/p+c/pan, so bn =1/an.
If b≠0, let an+ 1+x=y(an+x)/qan+c, compare with the known recurrence formula to get x and y, let bn=an+x, and get bn+ 1=ybn/qan+c, which is converted into b=0.
Type five
The recurrence formula is an+ 1/an=qn/n+k(q≠0, k∈N).
We can multiply the two sides of the equation (n+k) an+ 1=qnan by (n+k-1) (n+k-2) ... (n+1) to get (n+k-/kloc-). 6? 1nan, then bn+1= (n+k) (n+k-1) (n+k-2) ... (n+1) an+1.
So bn+ 1=qbn, so the sequence {bn} is the common ratio q, and the first term is b1= k (k-1) (k-2) ... 6? 1 1? 6? 1a 1=k！ The geometric series of a 1, and then an can be obtained.
In a word, the problem of finding the general term formula from the recursive formula of sequence is complicated and it is impossible to discuss them one by one. But as long as we grasp the recursive relationship of recursive sequence, analyze the structural characteristics and be good at reasonable deformation, we can find an effective way to solve the problem.
Type one? Inductive conjecture proof
We can write the first few terms of the sequence from the recursive formula of the sequence, then sum up the rules from the first few terms, guess a general term formula of the sequence, and finally prove it by mathematical induction.
? Example 1? Let the series {an} be a positive series with the first term 1, (n+1) a2n+1-nan2+an+1an = 0 (n =1,2, 3, ...
Solution: decompose (n+1) a2n+1-nan2+an+1an = 0 (n =1,2,3, ...) into (an+1+an).
Because an > 0, (n+ 1)an+ 1=nan, that is, an+1= n/(n+1) an.
Therefore, A2 = (1/2) a1= (1/2), A3 = (2/3) A2 = (1/3), ... Guess an = (1/).
Type two? "Difference Law" and "Product Commercial Law"
(1) When the recurrence formula of a series can be changed into an+ 1-an=f(n), take n = 1, 2, 3, ..., n- 1, and get n- 1 formulas.
a2-a 1=f( 1)，a3-a2=f(2)，…，an-an- 1=f(n- 1)，
And if f (1)+f (2)+…+f (n-1) can be found, the general term an can be accumulated on both sides. This method is called "step by step method".
Example 2? It is known that the sequence {an} satisfies a 1 = 1, an = 3n- 1+an- 1 (n ≥ 2). It is proved that an = (3n- 1)/2.
(Question 19 in the 2003 National Mathematics Volume)
It is proved that from the known an-an- 1=3n- 1, therefore,
an =(an-an- 1)+(an- 1-an-2)+…+(a2-a 1)+a 1 = 3n- 1+3n-2？ +…+3+ 1=3n- 1/2。
So I got the certificate.
(2) When the recurrence formula of a series can be changed into an+ 1/an=f(n), let n= 1, 2,3, …, n- 1, then we can get n- 1 formulas, that is.
a2/a 1=f( 1)，a3/a2=f(2)，a4/a3=f(3)，…，an？ /an- 1？ =f(n- 1)？ ,? If f (1) f (2) f (3) ... you can get f (n- 1), and you can get an by multiplying the two sides. This method is called product management method.
Example 3? (The same example is1) (Question 15 of National Mathematics Volume in 2000)
Another solution: (n+1) a2n+1-nan2+an+1an = 0 (n? = 1, 2, 3, ...) and get (n+ 1)an+ 1=nan, that is.
an+ 1/an=n/(n+ 1)。 ?
So an=an/an- 1? 6? 1an- 1/an-2？ 6? 1an-2/an-3？ 6? 1…? 6? 1a2/a 1？ =n- 1/n？ 6? 1n-2/n- 1？ 6? 1n-3/n-2？ 6? 1 … ? 6? 1 1/2? = 1/n。
Type three? structured approach
The recurrence formula is pan=qan- 1+f(n)(p and q are nonzero constants), and a new geometric series solution can be constructed by using the undetermined coefficient method.
Example 4? (Same Example 2) (Question 19 of the 2003 National Mathematics Volume)
Another solution: 3 is obtained from an=3n- 1+an- 1 6? 1an/3n = an- 1/3n- 1+ 1。
Let bn = an/3n, then there is
bn = 1/3bn- 1+ 1/3。 (*)
Let bn+x= 1/3(bn- 1+x), then bn =1/3bn-1+/3x-x, compared with formula (*), x =-60. Therefore, bn-1/2 =1/3 (bn-1-1/2). So the sequence {BN- 1/2} is the first term of B 1- 1 = A. 6? 1( 1/3)n- 1, which means an/3n-1/2 =-1/6 (1/3) n-1. Therefore, an = 3n.
Example 5? In the sequence {an}, a 1 = 1, an+ 1 = 4an+3n+ 1, find an.
Solution: let an+1+(n+1) x+y = 4 (an+NX+y), then
An+ 1 = 4an+3x+3y-x, compared with the known an+ 1=4an+3n+ 1, we get
3x=3, so
x= 1，
3y-x= 1，y=(2/3)。
So the sequence {an+n+(2/3)} is a geometric series with the first term a 1+ 1+(2/3)=(8/3), so an+n+(2/3)=(8/3)? 6? 14n- 1, i.e.
Ann =(8/3)? 6? 14n- 1-n-(2/3)。
Another solution: when n≥2 is known, an = 4an-1+3 (n-1)+1,which is different from the known relationship, an+1-an = 4 (an-an-1. Therefore, the sequence {an+ 1-an+ 1} is a geometric series with the first term a2-a1+1= 8, and the general term an can be obtained by "differential method". 6? 14n- 1-n-(2/3)。
Type four? Can be converted into
Type 3 general terms
(1) "logarithmic method" is converted into
The third kind.
The recurrence formula is an+ 1=qan? K (q > 0, k ≠ 0 and k ≠ 1, a 1 > 0), take the common logarithm on both sides and get lgan+ 1=klgan+lgq. Let lgan=bn, then we have bn+/kloc-.
The third kind.
Example 6? Given the sequence {an}, a 1=2, an+ 1=an2, find an.
Solution: from an+ 1 = AN2 > 0, the logarithm of both sides is lgan+ 1 = 2lgan. Let bn=lgan, then bn+ 1 = 2bn. Therefore, the first term of the sequence {bn} is b 1 = lga 1 =.
(2) Convert the "reciprocity method" into
The third kind.
The recurrence formula is in the form of quotient: an+1= (pan+b)/(qan+c) (an ≠ 0, PQ ≠ 0, PC ≠ QB).
If b=0, an+ 1 = pan/(qan+c). Because an≠0, the reciprocal of both sides is 1/an+ 1=q/p+c/pan, so bn =1/an.
The third kind.
If b≠0, let an+ 1+x=y(an+x)/qan+c, compare with the known recurrence formula to get x and y, let bn=an+x, and get bn+ 1=ybn/qan+c, which is converted into b=0.
Example 7? In the sequence {an}, a 1 = 2, an+1= (3an+1)/(an+3) are known, and the general term an is found.
Solution: Let an+ 1+x=y(an+x)/an+3, then an+1= (y-x) an+(y-3) x/an+3, combined with the known recurrence formula.
Y-x=3, so
x= 1，
y-3= 1，y=4，
Then an+1+1= 4 (an+1)/an+3, and bn=an+ 1, then bn+ 1=4bn/bn+2, and the reciprocal is/kloc-0. 6? 1 1/bn+ 1/4, i.e.1/bn+1/2 =1/2 (1/bn-65438+
Therefore, the first term of the sequence {1/BN- 1/2} is1/b1-1/2 =1/a1-65438.
So1/bn-1/2 = (-1/6) (1/2) n-1,you can get an.
Type five? The recurrence formula is an+ 1/an=qn/n+k(q≠0, k∈N).
We can multiply the two sides of the equation (n+k) an+ 1=qnan by (n+k-1) (n+k-2) ... (n+1) to get (n+k-/kloc-). 6? 1nan, then bn+1= (n+k) (n+k-1) (n+k-2) ... (n+1) an+1.
So bn+ 1=qbn, so the sequence {bn} is the common ratio q, and the first term is b1= k (k-1) (k-2) ... 6? 1 1? 6? 1a 1=k！ The geometric series of a 1, and then an can be obtained.
Example 8? (The same example is1) (Question 15 of National Mathematics Volume in 2000)
Another solution: simplify (n+1) a2n+1-Na2n+an+1an = 0 (n =1,2,3, ...) to (n+1). 6? 1a 1= 1, so bn= 1, that is, nan= 1, so an =1/n.
In a word, the problem of finding the general term formula from the recursive formula of sequence is complicated and it is impossible to discuss them one by one. But as long as we grasp the recursive relationship of recursive sequence, analyze the structural characteristics and be good at reasonable deformation, we can find an effective way to solve the problem.