The master quickly solves a series of problems known, a1 2, a n 1 4 an 3 an 2 , try to find the gene

Let bn=an+1 then an=bn-1

Substituting: b(n+1)-1=(4bn-4+3) (bn-1+2)=(4bn-1) (bn+1).

b(n+1)=(4bn-1)/(bn+1)+1=5bn/(bn+1)

To pour it down: 1 b(n+1) = (bn+1) 5bn = 1 5+1 5bn

That is, 1 b(n+1)-1 4=1 5(1 bn-1 4).

Let 1 bn-1 4=cn

Then c(n+1)=1 5cn c1=1 b1-1 4=1 (a1+1)-1 4=1 3-1 4=1 12

So cn is proportional.

cn=(1/12)*(1/5)^(n-1)

1/bn=cn+1/4

So bn=1 [(1 12)*(1 5) (n-1)+1 4].

an=bn-1=1/[(1/12)*(1/5)^(n-1)+1/4]-1

Simplify to obtain: an=(9*5 (n-1)-1) (3*5 (n-1)+1).

a(n+1)=(4*an+3) (an + 2)a(n+1)+1=5(an+1) (an + 2)1 [a(n+1)+1]=1 5*[1+1 (an+1)] Let bn=1 (an+1).

then b(n+1)=1 5*[1+bn].

b(n+1)-1 4=1 5(bn-1 4)bn-1 4 is a proportional series of 1 5.

b1-1/4=1/3-1/4=1/12

then bn-1 4=1 12*(1 5) (n-1) then an=1 [1 4+1 12*(1 5) (n-1)]-1

an+a(n+1)=4n+1

a(n-1)+an=4n-3

Subtract a(n+1)-a(n-1)=4

So a1, a3, a5 ......It is an equal difference guess hall column with a tolerance of 4.

A2, A4, A6 are also tolerance spike silver hidden for the difference series of the book 4.

an=3+a(n-1), so an-a(n-1) 3, the number series is a series of equal differences, and the tolerance is 3, because a1 5

So an=a1+(n-1)*d=5+3n-3=3n+2

an+1=(n+2) n*an, is a(n+1) an=(n+2) n, therefore, a2 ,n terms are multiplied by both sides of =3 The above two equations are equal, n(n+1) 2=an a1,a1=2 an=n(n+1).

N2, a(n+1)=(n-1)an (n-an)1 a(n+1)=(n-an) [n-1)an]=n [(n-1)an] -1 (n-1).

Both sides of the equation are divided by n

1 [na(n+1)]=1 [(n-1)an]-1 hail [n(n-1)]=1 [(n-1)an]-[1 (n-1)-1 n].

1 n)[1 a(n+1)-1]=[1 nuisance(n-1)][1 an -1].

1 (2-1)](1 a2 -1)=4-1=3 The sequence starts with the second term, each of which is equal to 3

1/(n-1)][1/an -1]=3

1/an =3n-2

an=1/(3n-2)

When n=1, a1=1 (3-2)=1, the same full source stove is full of prosperity.

The general formula for a series of numbers is an=1 (3n-2).

A n+2=4(a n+1)-4(an) is transferred to the end of the term, and a n+2 -2a n+1=2( a n+1 -2a n) can be seen a n+1 -2a n is a common ratio of 2, and the first term is the proportional number of the seepage a2-2a1=3. It is possible to find a n+1 -2a n =3*2 n-1 and divide both sides by 2 n at the same time, and get a n+1 2 n+1 - a n 2 n=3 4 and see an 2 n....

The 'n-1' in the description of '2an-1+1' in your question should be the subscript of the number series (n-1), help you solve it on this premise): first write the first few items of the number series according to the given entry 3+2 2+2+1、...The n terms of the number chain pie are 2 (n-1)+2 (n-2)+2 (n-3)+....2^2+2^1+2^0.

This is the sum of the first n terms of the proportional series, so the general term of the series is easily obtained according to the summation formula of the proportional series: an=(2 n)-1

The method of finding the formula of the grinding term in the middle of the beat:

1. Simplify the item an for each number of buckets

an = 2an-1 6an-1 + 1)2, using mathematical induction, the simplified formula is written in the form of a general formula:

an = 2^n * a1) /6^n * a1) +2^(n-1) +2^(n-2) +2^0))

3. Substitute a1 into the general formula:

an = 2^n * 1/7) /6^n * 1/7) +2^(n-1) +2^(n-2) +2^0))

The formula for getting the general term is:

an = 2^n * 1/7) /6^n * 1/7) +2^n - 1))

Note: Step 3 may need to be further demonstrated using mathematical induction.

a(n+1)=2an/(6an+1)

Countdown on both sides. 1/a(n+1)=(6an+1)/(2an)3 + 1/(2an)

1 a(n+1) +3 = 2( 1 an + 3)> is proportional to the collapse of the column, q=2

1/an + 3 = 2^(n-1).(1/a1 + 3)= 3 +

an = 1/(-3 +

The number series talks about the general formula of Jian An.

an=1/(-3 +

The deformation of this topic is actually not difficult, and the main thing is to see the relationship between the reciprocal.

The details are as follows:

Solution: A(n+1)=(n-1)an (n-an)1 a(n+1)=(n-an) [(n-1)an]=n [(n-1)an] -1 (n-1).

Both sides of the equation are divided by n

1/[na(n+1)]=1/[(n-1)an]-1/[n(n-1)]=1/[(n-1)an]-[1/(n-1)-1/n]

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

1 (2-1)](1 a2 -1)=4-1=3 The sequence starts with the second term, each of which is equal to 3

1/(n-1)][1/an -1]=3

1/an =3n-2

an=1/(3n-2)

n=1, a1=1 (3-2)=1, the same is satisfied.

The general formula for a series of numbers is an=1 (3n-2).