Known sequence an , a1 3, an 2 n a n 1 , the general formula for finding the sequence an

First, recursively prove that an is a non-zero sequence. (If an is a non-zero number, then a(n+1) is also a non-zero number, and a1 is a non-zero number).

The column equation is as follows.

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

a(n-1)=a(n-2)*2^(n-1)a2=a1*2*2 (n>=2)

Multiply, eliminate the common term, get an=3*2 (2+3+4+......n)=3*2 ((n+2)*(n-1) 2) The test n=1 also satisfies this equation, therefore.

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

Obtained by, an=2 n*a(n-1).

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

So a2 a1=2 2=4

a3/a2=8

a4/a3=16

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

Multiply the left and right sides of each equation to obtain an a1=4 8 16 ......2^n=2^[(n+2)(n-1)/2] n≥2

And a1=3 so an=3·2 [(n+2)(n-1) 2] n 2 When n=1, the above equation is also satisfied.

Therefore an=3·2 [(n+2)(n-1) 2] n 1

By accumulation, an-a(n-1)=2n-1 can be accumulated

a(n-i)-a(n-2)=2(n-1)-1a2-a1=2*2-1

So. an-a1=n^2-1

And because a1=2

So an=n 2+1

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

So a(n+1)-1=2an-1-1=2an-2=2(an-1), so the sequence is proportional to a1-1=3-1=2 and 2 is the common ratio.

So an-1=2*2 (n-1)=2 n

So an=2 n+1

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

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

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

1 a(n+1) -1] (1 an -1)=1 2, is a fixed value.

The series of numbers that come 1 a1 -1=1 (2 3) -1=3 2 -1=1 2 is a sequence of equal proportions du with 1 2 as the first term and 1 2 as the common ratio.

1/an -1=(1/2)×(1/2)^(n-1)=1/2ⁿ1/an=1+ 1/2ⁿ=(2ⁿ+1)/2ⁿan=2ⁿ/(2ⁿ+1)

When n=1, a1=2 (2+1)=2 3, which also satisfies the general term DAO

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

Answer: Because a(n+1)=2an (1+an), 1 a(n+1)=(1+an) (2an)=1 2+1 (2an).

Req =, when n = 1 b1 = 3 2

So b(n+1)=1 2+bn 2

With the undetermined coefficient method: b(n+1)+k=(bn+k) 2, that is, b(n+1)=(bn-k) 2, i.e., -k=1, so k=-1;

So b(n+1)-1=(bn-1) 2

i.e. [b(n+1)-1] (bn-1)=1 2, when n=1, b1-1=1 2

Therefore, it is a proportional series with the first term being 1 2 and the common ratio being 1 2.

So bn-1=1 2 n

So bn-1=1 2 n

So an=1 (1+1 2 n)=2 n (1+2 n)=1-1 (1+2 n).

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

Solution: an+1=2an (1+an), take the reciprocal to get: 1 (an+1)=(1+an) 2an, that is, 1 (an+1)=1 2+1 2an, minus 1 on the left and right sides

It gets: 1 (an+1) -1=1 2an-1 2=(1 an-1) 2, that is: [1 (an+1) -1] [(1 an-1)]=1 2, so the sequence is a proportional sequence with the common ratio of 1 2 and the prime phase of 1 2, 1 an-1=(1 2)*(1 2) (n-1)=(1 2) n, so an=1 [(1 2) n+1].

Your question is yes. Can't answer. I don't even know if your an+1 is n+1 or (an)+1

a1=2

a2-a1=3 yields: a2=5

a3-a2=6 gets: a3=11

a4-a3=9 gets: a4=20

Their difference is a multiple of 3, from which we get:

an=2+3n(n-1)/2

a（n+1）=a（n）+3n

a（n）=a（n-1）+3（n-1）

a（n+1）=a（n-1）+3【（n-1）+n】……Get: a(n)=a(1)+3 [1+2+3+......n-1）】=2+【3n（n-1）】/2

This is the process of solving the question. Fill-in-the-blank questions or multiple-choice questions. You know.

a2-a1=3

a3-a2=6

a4-a3=9

an-an-1=3(n-1)

Adding all of the above equations gives an-a1=3+6+9+......The right side of the equation 3 (n-1) uses the equal difference series, and the left side is known, you can find it, I hope it will help you.

Answer: From a(n+1)=nan (n+1):

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

So: a2 a1=1 2;

a3/a2=2/3

a4/a3=3/4

an a(n-1) = (n-1) n (where n 2) is multiplied to give :

an/a1=1/n

So an=a1 n=2 3n

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

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

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

So {1 an-1} is a proportional sequence with the first term -1 2 and the common ratio is 1 2, so 1 an-1=-(1 2) n

So an=1 [1-(1 2) n]=2 n (2 n-1).

2)ai(ai-1)=2^i/(2^i-1)^2=1/(2^i+1/2^i-2)

Since a1(a1-1)+a2(a2-1)=2+4 9=22 9

And when i>=3, ai(ai-1)=2 i (2 i-1) 2=1 (2 i+1 2 i-2)<=1 (2 i-2)<=1 2 (i-1).

So (i=3 to n) ai(ai-1)<=1 4+1 8+...1/2^(n-1)<=1/2

Therefore, (i=1 to n) ai(ai-1)<=22 9+1 2=53 18<3

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

Left and right are counts down. 1/a(n+1)=(an+1)/2an=1/2+1/2an

So (I wonder if you have copied a 2 less on the topic) let me know if you don't. I keep doing it.

1) Let bn=1 an, then since b1=1 2, because a(n+1)=2an (an+1), so b(n+1)=bn 2+1 2, i.e. b(n+1)-1=(bn-1) 2

Therefore, it is a proportional series with -1 2 as the first term and 1 2 as the common ratio, bn-1=-(1 2) n

So bn-1=1 2 n

i.e. an=1 bn=2 n (2n-1).

2)ai(ai-1)=2^i/(2^i-1)^2=1/(2^i+1/2^i-2)

Since a1(a1-1)+a2(a2-1)=2+4 9=22 9

And when i>=3, ai(ai-1)=2 i (2 i-1) 2=1 (2 i+1 2 i-2)<=1 (2 i-2)<=1 2 (i-1).

So (i=3 to n) ai(ai-1)<=1 4+1 8+...1/2^(n-1)<=1/2

Therefore, (i=1 to n) ai(ai-1)<=22 9+1 2=53 18<3

For a(n+1)=2an (an+1), both sides are taken from the reciprocal.

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

Multiply both sides of the above equation by 2 n to get 2 (n+1) a(n+1)=2 n an+2 n

Let b(n+1)=2 (n+1) a(n+1), then bn=2 n an then has: b(n+1)-bn=2 n, b1=2 1 a1=1

So: bn-b(n-1)=2 (n-1).

b(n-1)-b(n-2)=2^(n-2)

b2-b1=2

According to the accumulation method: bn-b1=2+......2^(n-2)+2^(n-1)

So bn = 1 + 2 + ......2^(n-2)+2^(n-1)=1(1-2^n)/(1-2)=2^n-1

i.e. 2 n an=2 n-1 then an=2 n (2 n-1).

ai（ai--1）=2^n/(2^n-1)[2^n/(2^n-1)-1]=2^n/(2^n-1)[1/(2^n-1)]=2^n/(2^n-1)^2>0

Reciprocal on both sides: 1 [ai(ai--1)]=(2 n-1) 2 2 n=[2 (2n)-2*2 n+1] 2 n=2 n+1 2 n-2

Then the sum is reduced according to the proportional sequence.