Cumulative method
The recurrence formula is a(n+ 1)=an+f(n), and f(n) can be summed.
Example: the sequence {an} satisfies the general formula of a1=12, and a (n+1) = an+1(4N2-1) finds {an}.
Solution: A (n+1) = an+1/(4N2-1) = an+[1(2n-1)-1(2n+60)
∴an=a 1+( 1- 1/3+ 1/3- 1/5+……+ 1/(2n-3)- 1/(2n- 1))
∴an= 1/2+ 1/2( 1- 1/(2n- 1))=(4n-3)/(4n-2)
Cumulative multiplication
The recurrence formula is a(n+ 1)/an=f(n), and f(n) can be integrated.
Example: the sequence {an} satisfies a(n+ 1)=(n+2)/n an, and a 1=4, so find an.
Solution: an/a1= an/a (n-1) × a (n-1)/a (n-2) × …× a2/a1= 2n (n+/kloc-0
structured approach
Transform non-arithmetic progression and geometric progression into relevant arithmetic progression.
Properly carry out operational deformation.
Example: in {an}, a 1 = 3, a (n+ 1) = an 2, and find an.
Solution: LNA (n+ 1) = LNAn 2 = 2LNAn.
∴{ln an} is a geometric series, q=2, and the first term is ln3.
∴ln and = (2 (n-1)) ln3
Therefore, an = 3 [2 (n- 1)].
Reciprocal transformation method (applicable to a(n+ 1)=Aan/(Ban+C), where a, b, C∈R)
Example: in {an}, a 1= 1, a(n+ 1)=an/(2an+ 1).
Solution:1/a (n+1) = (2an+1)/an =1/an+2.
∴{ 1/an} is arithmetic progression, the first term is 1, and the tolerance is 2.
∴an= 1/(2n- 1)
method of undetermined coefficients
A. the recurrence formula is a(n+ 1)=pan+q(p, q is constant), and the recurrence sequence {an+x} can be constructed as a geometric series with p as the common ratio.
That is, a(n+ 1)+x=p(an+x), where x=q/(p- 1) (or the set formula can be decomposed and equal to atoms).
Example: in {an}, a 1= 1, a(n+ 1)=3an+4, find an.
Solution: a(n+ 1)+2=3(an+2)
∴{an+2} is a geometric series, the first term is 3 and the common ratio is 3.
∴an=3^n-2
The recurrence formula of b is a (n+ 1) = pan+q n (p, q is constant).
Conventional deformation, with both sides divided by q (n+ 1),
Get a (n+1)/q (n+1) = p/q an/q n+1/q.
Let bn = an/q n,
You can get b (n+ 1) = kbn+m (k = p/q, m = 1/q).
Then use the method mentioned in a above to solve it.
The recurrence formula of c is a(n+2)=pa(n+ 1)+qan, (p and q are constants).
Let a (n+2) = x 2, a (n+ 1) = x, an = 1.
Two formulas can be obtained by solving x 1 and x2.
a(n+ 1)-x 1an = x2(an-x 1a(n- 1))
a(n+ 1)-x2an = x 1(an-x2a(n- 1))
Then subtract the two expressions, and kan can be obtained on the left.
On the right, you can get it by geometric series.
Example: in {an}, a 1 = 1, a2 = 2, a (n+2) = 2/3a (n+1) =1/3an.
Solution: x 2 = 2x/3 = 1/3.
x 1= 1，x2=- 1/3
You can get the equation.
a(n+ 1)-an =- 1/3(an-a(n- 1))
a(n+ 1)+ 1/3 an = an+ 1/3 a(n- 1)
The solution is an = 7/4-3/4× (-1/3) (n-1).
D recurrence formula a(n+ 1)=pan+an+b(a, b and p are constants).
It can be converted into a (n+1)+x (n+1)+y = p (an+xn+y).
Then compared with the original formula, we can get x, y,
That is to say, {an+xn+y} is a geometric series with p as the common ratio.
Example: in {an}, a 1=4, an = 3a (n-1)+2n-1(n ≥ 2).
Solution: original formula = & gtan+n+1= 3 [a (n-1)+(n-1)+1]
∴{an+n+ 1} is a geometric series, q=3, and the first term is 6.
∴an=2×3^n-n- 1
Characteristic root method
The recurrence formula is a (n+1) = (aan+b)/(can+d) (a, b, c and d are constants).
Let a(n+ 1)=an=x, and the original formula is x=(Ax+B)/(Cx+D).
(1) If the same real root x0 is solved, the sequence {1/(an-x0)} can be constructed as a arithmetic progression.
Example: {an} satisfies a 1=2, and a (n+1) = (2an-1)/(4an+6) to find an.
Solution: x=(2x- 1)/(4x+6)
X0=- 1/2。
1/(an+ 1/2)= 1/[(2a(n- 1)- 1)/(4a(n- 1)+6)+ 1/2]= 1/[a(n- 1)+ 1
∴{ 1/(an+ 1/2)} is arithmetic progression, d= 1, and the first term is 2/5.
∴an=5/(5n-3) - 1/2
(2) If two different real roots x 1, x2 are solved, the structure {(an-x 1)/(an-x2)} is a geometric series (x 1, x2 is dislocated and interchangeable).
Example: {an} satisfies a 1=2, and a (n+1) = (an+2)/(2an+1).
Solution: We can get (an-1)/(an+1) =-1/3 [a (n-1)-1]/(a (n-/kloc-0
Then {(an- 1)/(an+ 1)} is a geometric series, q=- 1/3, and 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, then this series may be a periodic series.
Example: in {an}, a 1=2, which satisfies a (n+1) = (an-1)/an (n ≥ 2).
Solution: A 1 = 2, A2 = 1/2, A3 =- 1, A4 = 2, A5 = 1/2. ...
So an=2(n MOD 3= 1), 1/2(n MOD3= 1),-1(nMOD3=0).
(To be exact, there should be a piecewise function such as braces, but it can't be shown here. )
Addition and subtraction, multiplication and division
Example: {an} satisfies a1+2a2+3a3+...+nan = n (n+1) (n+2).
Solution: Let BN = a1+2a2+3a3+...+nan = n (n+1) (n+2).
nan = bn-b(n- 1)= n(n+ 1)(n+2)-(n- 1)n(n+ 1)
∴an=3(n+ 1)
General formula: A series of numbers arranged in a certain order is called a series, and the nth item {a n} of the series is expressed by a specific formula (including parameter n), which is called the general formula of the series. This is like the analytic expression of a function. By substituting a specific value of n, the corresponding value of A-n can be obtained. The solution of the general term formula of sequence is usually obtained by transforming its recursive formula many times.