The general formula of a second-order sequence

The second-order sequence of the general formula of the sequence

According to the concept of a first-order recursive sequence, we can define a recursive expression that simultaneously contains an 2, an 1, and an as a second-order sequence. Compared with the first-order sequence, the general term formula of the second-order sequence is more complicated. In order to facilitate the transformation, let us first explain the simple form of the second-order sequence:

an 2 = A * an 1 B * an , (Similarly, A and B are constant coefficients) The basic idea is similar to the first order, but when compounding, pay attention to the undetermined coefficients and corresponding terms

Composition of the original formula: Let the original formula be transformed into this form an 2 - ψ * an 1 = ω (an 1 - ψ * an)

Compare this formula with the original formula, we can get

ψω = A and -(ψ*ω) = B

The values of ψ and ω can be obtained by solving these two equations,

Let bn = an 1 - ψ*an, the original formula becomes bn 1 = ω *bn geometric sequence, and the bn general term formula bn= f (n) can be obtained,

Through the given equation an 1 - ψ*an = f(n), we can observe that this formula is actually the definition of a first-order sequence. This formula only involves two sequence variables an 1 and an, so it can be regarded as "order reduction", converting a second-order sequence into a first-order sequence to solve the problem.

It is known that the general term of the quadratic second-order recursion formula of a certain sequence

A(n 1)=A(n) A(n-1)-2A(n)*A(n-1)

Deformed into 1-A(n 1)=(1-An)(1-A(n-1))

Let Bn=1-An, get

B(n 1)=Bn*B(n-1)

If it can be guaranteed that Bn>0, then you can take the logarithm of both sides to get lgB(n 1)=lgBn lgB(n-1)

Then let Cn=lgB(n 1), then Cn becomes the Fibonacci sequence, which is omitted below

If Bn>0 cannot be guaranteed, observe B3=B2B1

B4=(B2)^2*B1

B5=(B2)^3*(B1)^2

B6=(B2)^5*(B1)^3

Note that Bn=(B2)^x*(B1)^y

Obviously x and y are both Fibonacci numbers, the following will be omitted

(For the Fibonacci sequence, you can search online. Its general terms are more complicated and are not written here)

Note that the result obtained by using the above method may be Cn or Bn, and you need to convert An=1-Bn at the end. Don’t forget it

How to derive the general term formula from the second-order recursion formula?

a(n 1) pan qa(n-1)=0

Suppose a(n 1) xan=y[an xa(n-1)]

a(n 1) (x-y)an-xya(n-1)=0

x-y=p

xy=-q

x1=p√（p^2-4q）,y1=√（p^2-4q）,

x2=p-√（p^2-4q）,y2=-√（p^2-4q）,

a(n 1) x1an=y1[an x1a(n-1)]

a(n 1) x2an=y2[an x2a(n-1)]

Division of two equations:

[a(n 1) x1an]/[a(n 1) x2an]=(y1/y2){[an x1a(n-1)]/[an x2a(n-1)]}

Suppose bn=[a(n 1) x1an]/[a(n 1) x2an]

bn=(y1/y2)b(n-1)=-b(n-1)

bn=b1(-1)^(n-1),b1=[a2 x1a1]/[a2 x2a1]

[a(n 1) x1an]/[a(n 1) x2an]=b1(-1)^(n-1)

a(n 1) x1an=b1[a(n 1) x2an](-1)^(n-1)

=[b1(-1)^(n-1)]a(n 1) [b1(-1)^(n-1)]x2an

[1-b1(-1)^(n-1)]a(n 1)={[b1(-1)^(n-1)]x2-x1}an

[1-b1(-1)^(n-2)]an={[b1(-1)^(n-2)]x2-x1}a(n-1)

[1-b1(-1)^(n-3)]a(n-1)={[b1(-1)^(n-3)]x2-x1}a(n-2)

……

[1-b1(-1)^2]a4={[b1(-1)^2]x2-x1}a3

[1-b1(-1)^1]a3={[b1(-1)^1]x2-x1}a2

[1-b1(-1)^0]a2={[b1(-1)^0]x2-x1}a1

Multiply both sides:

[1-b1(-1)^(n-2)][1-b1(-1)^(n-3)]……[1-b1(-1)^2][1-b1 (-1)^1][1-b1(-1)^0]an

={[b1(-1)^(n-2)]x2-x1}{[b1(-1)^(n-3)]x2-x1}……{[b1(-1)^ 2]x2-x1}{[b1(-1)^1]x2-x1}{[b1(-1)^0]x2-x1}a1

The coefficients on both sides are known, and an is out (as long as a1 is provided).

If p and q are specific numbers, both sides can be simplified.

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