In the sequence a n, a1 1, a n 1 2a n 2 a n. Ask a2 a3 a4 The second question is an process

Is this the question: a(n+1)=(2an) (2+an) yes, the answer is as follows;

a2=2/3 a3=1/2 a4=2/5

1 a(n+1) =(2+an) 2an =1 an +1 2 Let bn=1 an then b1=1 b(n+1)=bn+1 2bn=(n+1) 2 (n 2) and when n=1, b1=(1+1) 2=1 is true.

bn=(n+1) 2 an=2 (n+1)Supplement: For a series of numbers of the form an=ba(n-1) [ca(n-1)+d], you can take the reciprocal to create an equal difference series.

One bridge imitation counts:

an+a(n+2)+an*a(n+2)=1；a1=1/2,a2=1/4；

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

a3=(1-a1)/(1+a1)=(1-1/2)/(1+1/2)=1/3;

a4=(1-a2)/(1+a2)=(1-1/4)/(1+1/4)=3/5;

a5=(1-a3) (1+a3)=(1-1 3) (1+1 3)=1 Burning blindness 2;

a6=(1-a4)/(1+a4)=(1-3/5)(1+3/5)=1/4;

a5+a6=1 Min Duan fiber 2+1 4=3 4

a(n+1)/an=2

So it is the first proportional series with an average ratio of 1 to 2.

So a2=2

a2,a4,a6……The number series composed of a2n is a proportional series with the first term being 2 common and 4.

Therefore, the substitute side shouts the power of the fiber and the formula can erect the oak .........I forgot the formula.

Calculate it yourself, and use the dislocation subtraction method to calculate Zheng Xingxia,

When n=2, an=a2=2 2 a1=4

When n>2 and n-1 is greater than or equal to 2, there is.

a1*a2*……a(n-1)=(n-1)^2...1

a1*a2*……an=n^2...Divide 22 by 1.

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

Apparently when n=2 epoch into is still true.

So an=n 2 (n-1) 2(n>=2)a4=4 2 (4-1) 2

a4=16/9

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

a1a2a3••。an= n²

Count one less place, and replace n with n-1;

a1a2a3••。an-1= (n-1) is a one-way comparison.

a1a2a3••。an= n ·· Substituting n with (n-1) yields:

a1a2a3••。an-1= (n-1) · Use @ to get: an=n (n -1).

Thank you for adopting...

a2 = 2 3, a3 = 1 2, a4 = 2 5, a5 = 1 3 take "an=2an-1 2+an-1" as the reciprocal number.

1/an=1/an-1+1/2

This shows that the sequence s is a proportional series with 1 as the first term and 1 2 as the common ratio.

Further reasoning, yes.

an=2/(n+1)

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

Take the reciprocal number. 1/a(n+1)=(a(n)+1)/2a(n)[1/a(n+1)-1]=1/2[1/an -1]1/a1-1=-1/2

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

1/an=1-（1/2）^n

an=1 [1-(1 2) n] n is a positive integer 2) so that bn=an*(an-1)=1 [1-(1 2) n] *1 2) n [1-(1 2) n].

1/2）^n/[1-（1/2）^n]^2=2^n/(2^n-1)^2<2^n*2/(2^(n-1))^2=2^n*2/2^2（n-1）=1/2^（n-1）b1=2

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

Sum of 3 proportional sequences.

Hello landlord.

It can be obtained from a(n+2)=4a(n+1)-3an, a(n+2)-a(n+1)=3(a(n+1)-an). Now let a(n+1)-an=bn, then b(n+1)=3bn, and b1=a2-a1=1, so bn=3 (n-1), i.e., an-a(n-1)=3 (n-2)......a2-a1=3 0, which adds up to an-a1=3 0+......3 (n-2)=1(1-3 (n-1)) (1-3)=(3 (n-1)-1) 2, so an=a1+(3 (n-1)-1) 2=1+(3 (n-1)-1) 2

Hope you're satisfied.

a1*a2*a3*……an=n^2 (1)a1*a2*a3*……an-1=(n-1) 2 (2)1) divided by (2) gives us an=[n (n-1)] 2a3=(3 2) 2=9 4

a5=(5 4) 2=25 16

a3+a6=61/16