a1 1,a2 1,a3 1, an a n 1 a n 2 a n 3 how to find his general formula

I'll do this for you, and I'll tell you how it works.

This problem belongs to the problem in "Combinatorics", and the knowledge required is to use the occurrence function to solve the linear homogeneous recursive relation of the constant coefficient. First, Definition 1:

The sequence satisfies: a(n) +b1*a(n-1) +bk*a(n-k) = 0,b1,..bk is a constant, so this relation is called a linear homogeneous recursive relation with constant coefficients.

Definition 2: The polynomial g(t) = t k + b1*t (k-1) + bk is the characteristic polynomial of the above recursive relation. It can be found that the eigenpolynomial is obtained by multiplying the recursive a(n-k) = 1 by t.

Theorem: If the eigenpolynomial corresponding to a recursive relation has k different roots x1, x2, .,xk ；Then the general term can be expressed as:

a(n) = c1 * x1 n + ck * xk n, where c1 ,..ck is taken over any plural number.

For your problem, the eigenpolynomial is g(t) = t 3 - t 2 - t - 1, let its 3 roots be x1, x2, x3; Then according to the theorem above, a(n) = c1 * x1 n + c2 * x2 n + c3 * x3 n. A(1) = 1, a(2) = 1, a(3) = 1;Then you can create a system of equations to solve c1, c2, c3. I'll help you do the math, it's more complicated, the root of the eigenpolynomial has two complex roots, and the coefficients c1, c2, and c3 are also complex.

I'll post screenshots of them:

If you think about it, when n=4, a4=a3+a2+a1=3, n=5, a5=a4+a3+a2=5

It is neither proportional nor equal difference, **There is a general formula. -

Anyway, I don't count. Let's see if there are any masters who can solve it.

The general term has a plural number of i

If you're sure you need it.

Can be done for you.

a[n+1]=3a[n]-2^n

Let b[n]=a[n]-2 n

Then b[n+1]=a[n+1]-2 (n+1)3a[n]-2 n-2 (n+1)=3(a[n]-2 n)b[n] is an equal proportional series, which can be imitated by b[n]=k3 n, substituted by b[1]=3-2=1, and the solution is b[n]=3 (n-1).

So the infiltration group a[n]=b[n]+2 n=3 (n-1)+2 n

Add 1 to both sides of the grip

a(n+1)+1=3(an+1)

So an+1 is the proportional kippi Zheng column, q=3

So an+1=(a1+1)*q (n-1)=2*3 (n-1)an=-1+2*3 (n-1).

Answer: a1=2a(n+1)=2(an) 2+1>0a(n+1) +1= 2* [an) 2 +1 ][a(n+1) +1 ] an) 2 +1 ]=2So: is the equal crack gap ratio series set bn= [a(n+1) +1 ] branch letter [(an) 2 +1 ]a2=2(a1) 2+1=2*4+1=9 then b1=(9+1) 4+.

Let bn=an-1

a(n+1)=(3an+1) (an+3)a(n+1)-1=(3an+1) (an+3)-1b(n+1)=2bn (bn+4) (pour it quietly) make fn=1 bn

f（n+1）=2fn+1/2

f（n+1)+1/2=2(fn+1/2)

fn+1 2 is a proportional series. Prepared.

fn+1/2=3·2^（n-2）

So. an=1/[3·2^（n-2）-1/2]+1

by a(n+1)=3an+2 i.e.

a(n+1)+1=3(an+1)

A sequence is a proportional series, the first term of which is a1+1=2, and the common ratio is 3 an+1=2*3 (n-1).

an=2*3^(n-1)-1

a(n+1)=3an+2

a(n+1)+1=3an+3＝3(an+1)[a(n+1)+1]/(an+1)=3

So it's a proportional series with a common ratio of 3!

an+1=(a1+1)[1-3 n] (1-3)=3 n-1So: an=3 n-2

Solution: a(n+1)=3an+2 (n+1).

a(n+1)+2^(n+2)=3an+2^(n+1)+2^(n+2)=3an+3×2^(n+1)=3[an +2^(n+1)]

a(n+1)+2 (n+2)] [an+2 (n+1)]=3, which is a fixed value.

a1+2²=1+4=5

The number series is a proportional series with 5 as the first term and 3 as the common ratio.

The general formula for the series of an +2 (n+1)=5 3 (n-1)an=5 3 (n-1) -2 (n+1) is an=5 3 (n-1) -2 (n+1).

a(n+1)=2an+1

a(n+1)+1=2(an+1)

It is explained that the number hail Hu dismantling column is a proportional number series with a1+1 as the first source jujube term and 2 as the common ratio.

Old. an+1=(a1+1)*2^(n-1)=2^(n+1)an=2^(n+1)-1

a(n+1)=2an+1

a(n+1)+1=2（an+1 ）

Therefore, the number series is a proportional series with a1+1 as the first term, and 2 as the ratio of the public fiber.

an+1=(a1+1)*2 (n-1)=4*2 (n-1)=2 n, so the formula for Youhuitong term: an=(2 n)-1

a（n+1）=an/（2an+1）

Take the reciprocal on both sides:

1/a(n+1)=(2an+1)/an=2+1/an∴1/a(n+1)-1/an=2

is an equal difference series with a tolerance of 2

a1=1,1/a1=1

1/an=1/a1+(n-1)d

1+(n-1)*2

2n-1an=1/(2n-1)

For a(n+1)=an (2an+1), take the reciprocal count on both sides at the same time, 1 a(n+1)=2+ 1 an, so the series is a series of equal differences with 1 as the first term and 2 as the tolerance, and the general term is 1 an =2n-1So the formula for the general term of the number series is an=1 (2n-1).