It is known that the sequences an and bn satisfy a1 2, a2 4, and bn an 1 an,

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

b(n+1)+2=2[b(n)+2]

b(n)+2} is the first proportional sequence with b(1)+2=4 and a common ratio of 2.

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

a(n+1)=a(n)+2^(n+1) -2a(n+1)-2^(n+1+1)+2(n+1)=a(n)-2^(n+1)+2n=...=a(1)-2^2+2=0

a(n)=2^(n+1)-2n

bn+1 +2)=2(bn+2), b1 = 2, so bn = 2 (n+1).

bn/2=2^n

An = A1 + B1 + B2 + B3 + bn-1, so An 2 = 1+2+4+8+.2 (n-1) = 2 n-1, so an = 2 n

Finished. It should have turned out to be right......

There is a condition that is not clearly written: bn+1=2bn+2, you want to express b(n+1)=2b(n+2) [the one in parentheses is the subscript].

Do you still want to express b(n)+1=2b(n+2) or b(n+1)=2b(n)+2???Please write this condition clearly.

bn+1 +2=2(bn +2), so {bn} is a proportional series of common ratio q=2, because b1=a2-a1=4-2=2 so bn = 2 (n+1), so bn=2*2 n=2 (n+1), let's look at bn=a(n+1)-an, we think of accumulation: b1=a2-a1 b2=a3-a2 b3=a4-a3.........bn=a（n+1）-an

（1）b1=3/4

b2=b1 1-a1 2=4 5 so a2=1 5

b3=b2 1-a2 2=5 6 so a3=1 6

b4=b3/1-a3^2=6/7

2) Substituting bn=1-an into the second formula gives bn+1=(1-an) 1-an 2=1 (1+an).

1 (b(n+1)-1)- 1 (bn-1) = (-an+1) an)-(1 an)=-1 constant.

So the series is a series of equal differences.

3) 1 (bn-1) =-4-(n-1)= -(n+3) from (2).

So bn=(n+2) (n+3).

an=1/（n+3）

sn=(1/4)(1/5)+(1/5)(1/6)+(1/6)(1/7)+…1/(n+3))(1/(n+4))

1/4)-(1/5)+(1/5)-(1/6)+(1/6)-(1/7)+…1/(n+3))-1/(n+4))

1/4)-(1/(n+4))

n/[4(n+4)]

4asn=na/(n+4)

A<(n+2)(n+4) [(n+3)n] is obtained from 4asn bn

The limit of the formula n+2)(n+4) [(n+3)n] is 1

So a=1

a1=1/4，an+bn=1，b(n+1)=bn/(1-an^2)

b1=1-1/4=3/4

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

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

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

b(n+1)-1]/b(n+1)=bn-1

b(n+1)/[b(n+1)-1]=1/(bn-1)

b(n+1)/[b(n+1)-1]-1=1/(bn-1)-1

1/[b(n+1)-1]=1/(bn-1)-1

1/[b(n+1)-1]-1/(bn-1)=-1

So 1 (bn-1) is the first series of equal differences with an term of 1 (b1-1) = -4 and a tolerance of -1.

2 From above, we get 1 (bn-1)=-4-(n-1)=-n-3

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

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

sn=a1a2+a2a3+a3a4+…+ana(n+1)

1/4)(1/5)+(1/5)(1/6)+(1/6)(1/7)+…1/(n+2)][1/(n+3)]+1/(n+3)][1/(n+4)]

1/4-1/5)+(1/5-1/6)+(1/6-1/7)+…1/(n+2)-1/(n+3)]+1/(n+3)-1/(n+4)]

1/4-1/(n+4)

4asn bn again

4a[1/4-1/(n+4)]＜n+2)/(n+3)

n+3)a[(n+4)-4]＜(n+2)(n+4)

a-1)n^2+(3a-6)n-8＜0

If a-1 0, then y=(a-1)n 2+(3a-6)n-8 is a parabola with an opening upward, and there is always n so that (a-1)n 2+(3a-6)n-8 0, so it does not fit the topic;

If a-1 0, then y=(a-1)n 2+(3a-6)n-8 is the parabola with the opening downward, as long as 0, so.

3a-6) 2+32(a-1)=9a 2-4a+4 0, such a a does not exist;

If a-1=0, then y=(a-1)n 2+(3a-6)n-8=-3n-8 0 is constant;

In summary, a=1

The solution an=3 4a(n-1)+1 4b(n-1)+1(1)bn=1 4a(n-1)+3 4b(n-1)+1 (2)(1)+(2) gives an+bn=a(n-1)+b(n-1)+2,(n>=2)), so the sequence an+bn is an equal difference series with a1+b1=2 and a tolerance of 2.

cn=an+bn=2+(n-1)*2=2n

（1）b1=3/4

b2=b1/1-a1^2=4/5

So a2 = 1 5

b3=b2/1-a2^2=5/6

So a3=1 6

b4=b3/1-a3^2=6/7

2) Replace bn=1-an

Substitute the second formula.

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

1/(bn-1)

-(an+1) an)-(1 an)=-1 constant, so the series is a series of equal differences.

3) from (2).

1/(bn-1)

4-（n-1）=

n+3) so bn=(n+2) (n+3).

an=1/（n+3）

sn=(1/4)(1/5)+(1/5)(1/6)+(1/6)(1/7)+…1/(n+3))(1/(n+4))

1/4)-(1/5)+(1/5)-(1/6)+(1/6)-(1/7)+…1/(n+3))-1/(n+4))

1/4)-(1/(n+4))

n/[4(n+4)]

4asn=na/(n+4)

by 4asn bn

The limit of the formula a<(n+2)(n+4) [(n+3)n](n+2)(n+4) [(n+3)n] is 1, so a=1

b1=√a1a2=√2

b2=b1q=√a2a3,a3=b1^2q^2/a2=q^2

bn=b1q^(n-1)=√anan+1

bn+2=b1q^(n+1)=√an+1an+2

anan+1=2q^(n-1)

an+2an+1=2q^(n+1)

an/an+2=1/q^2

an+2=an *q^2

1. Certification. 2、cn=a(2n-1)+2a(2n)

a(2n+2)=q^2a(2n)

a(2n+1)=a(2n-1+2)=q^2a(2n-1)

cn+1/cn

a(2n-1+2)+2a(2n+2)]/[a(2n-1)+2a(2n)]

q^2*[a(2n-1)+2a(2n)]/[a(2n-1)+2a(2n)]

Q2 is a proportional series, and the common ratio is Q2

3、an+2=anq^2

1/a(2n)=1/a(2n-2+2)=1/q^2a(2n-2)=1/q^4a(2n-4)=1/q^6a(2n-6)

1/[q^2(n-1)a(2n-2n+2]

Similarly, 1 a(2n-1) = 1 q 2a(2n-3) = 1 q 2(n-1)a1

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

is the sum of two proportional series, the common ratio is q 2, and the first term is b1 = 1 a1 = 1, c1 = 1 a2 = 1 2

It's all n items. According to the summing formula:

s=(1-q^2n)/(1-q^2)+(1/2)(1-q^2n)/(1-q^2)

3/2)(1-q^n)(1+q^n)/(1-q)(1+q)

q≠±1q^2=1

then, a3=a1=a5=....=a(2n-1)=1

a2=a4=a6=...=a(2n)=1/2

s=n/2+n=3n/2

If you can't ask, I'll do my best to help you out

It is not easy to answer the question, if you are dissatisfied, please understand

an+bn=1

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

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

an-a(n+1)=ana(n+1)

Both sides of the equation are divided by ana(n+1).

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

1 a1=1 (1 4)=4, the series is a series of equal differences with 4 as the first term and 1 as the tolerance.

1/an=4+1×(n-1)=n+3

an=1/(n+3)

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

Mathematical induction, the first half of this question is the same, guess the item later, and then prove it, it's too troublesome, you can solve it yourself, but there are already the above results for comparison, it shouldn't be difficult.