It is known that the sequence an satisfies the condition 1 2a1 1 2 2a2 1 2nan 2n 5

Solution: (1) Let n=1 get 1 2a1=2*1+5=7, so a1=142)1 2a1+1 2*2a2+.1/2nan= 2n+5 (1)1/2a1+1/2*2a2+..

1 2(n-1)a(n-1) = 2(n-1)+5 (n greater than or equal to 2) (2).

1)-(2) gives an n=4 so an=4n (n is greater than or equal to 2) from (1) knows an=4n (n is greater than or equal to 2).

14 （n=1）

n 1, then 1 2a1 = 2*1 + 5 is obtained by the formula, and a1 1 14 is obtained;

n>1，an＝1/2a1+1/2*2a2+..1/2*(n-1)a(n-1)+1/2nan= 2n+5 (1)

a(n－1)＝1/2a1+1/2*2a2+..1/2*(n－1)*a(n－1)=2(n－1)+5 (2)

Using equation (1)-(2) to obtain 1 2*n*an=2 , then the general formula for an=1 4nan is an=1 14 (n=1)1 4n (n>1).

When n=1, 1 2a1=7 a1=14 1 2a1+1 2*2a2+.1/2*nan=2n+5 1/2a1+1/2*2a2+..1 2*nan+1 2*(n+1)a(n+1)=2(n+1)+5 subtract the two formulas to give 1 2*(n+1)a(n+1)=2 so a(n+1)=2*(n+2) so a1=7 when n is not equal to 1 an=2*(n+1).

With 1 2a1....1 2 nan+1 2 (n+1)an+1=2(n+1)+5 is used as the equation of 2 subtraction, and then the term is shifted.

1/2a1+1/2^2a2+..1/2^nan=2n+5……1 then replace an with an+1 to get it.

1/2a1+1/2^2a2+..1/2^nan+1/2^(n+1)a（n+1）=2(n+1)+5……22-1.

1/2^(n+1)a（n+1）=2

Solve an+1=2 (n+2).

Correspondingly, there is an=2 (n+1).

Or by the condition to know that the number series is a difference series, it is easy to find its general formula, and then find the an general formula.

Solution: Derived from the question.

an+1=2an+1

an+1)+1=2an+2

an+1)+1=2(an+1)

an+1)+1]/(an+1)=2

When n=1.

an+1=a1+1=2

When n=2.

Obtained by an+1=2an+1.

a2+1=2a1+1+1=4

Therefore, the number column is a proportional series with a common ratio of 2 and the first term is 2.

Proof: an+1+1=2an+2=2(an+1), and a1=1, a1+1=2≠0

Therefore, the number series is an equal proportional series with 2 as the first term and 2 as the common ratio

(1) Proof of: an an+1 (4an 2) (an+1 2) 4an 2) an (an+1 2) an+1

4 2 an 1 2 an+1 1 an+1 1 an 3 2 is a series of equal differences.

2)∵1/an+1－1/an＝3/2 ∴1/an＝1/a1＋(n－1)×3/2＝(3n＋2)/2 ∴an＝2/(3n＋2)

Suppose AK·AK+1 is the term AM

am＝ak·ak+1＝2/(3k＋2)·2/(3k＋5)＝4/[(3k＋2)·(3k＋5)]＝2/(3m＋2)

6m＋4＝9k²＋21k＋10 ∴m＝(3k²＋7k)/2＋1＝k(3k＋7)/2＋1

k(3k 7) is a multiple of 2 m n*

Existence ak·ak+1 is the term in and is the term k(3k 7) 2 1.

When n 1, and n z.

a1+1/（2²）*a2+……1/（2^n）*an=2n+5½a1+1/（2²）*a2+……1/（2^n）*an+1/（2^（n+1））*a（n+1）=2（n+1）+5

Subtract the two formulas to get.

1/（2^（n+1））*a（n+1）=2

a（n+1）=2*2^（n+1）

a（n+1）=2^（n+2）

When n 2, and n z, an=2 (n+1) when n=1, i.e., a1=2*1+5

a1=14So, the general formula for the series is an=14(n=1), an=2 (n+1)(n 2, n z).

These two steps are obviously recursively derived from the recursive and a2 values, and the other terms can be found according to the recursive formula.