Sn S n 1 An 2, A1 3, An 0, the general term formula An is known

sn+s（n-1）=an^2

s(n-1)-s(n-2)=a(n-1) 2 is subtracted by the formula, and the left side is an+an-1

The squared difference formula on the right (an-an-1) (an+an-1) is because an>0, so the two sides are reduced to an+an-1 to get an-an-1=1, so an is a series of equal differences with the first term being 3 and the tolerance of 1.

an=n+2

1、∵sn^2=a1^3+a2^3+…+an^3，sn-1^2=a1^3+a2^3+…+a(n-1) 3, subtract the two equations to get an 3=sn 2-s(n-1) 2=(sn-s(n-1)))sn+s(n-1)))=an(sn+s(n-1)),an 0, an 2=sn+s(n-1)(n 2),a(n-1 ) 2=s(n-1)+s(n-2()n 2), subtract the two formulas to get an2-an-12 =sn-s(n-2)=an+a(n-1), an-a(n-1)=1(n 3),s1 2=a1 2=a1 3,and a1 0, a1=1,s2 2=(a1+a2) 2=a1 3+a2 3,(1+a2) 2=1+a2 3, a2 3-a2 2-2a2=0, from a2 0, a2=2,an-a(n-1)=1,n 2, so the number column is an equal difference series, and the general formula is an=n

2、bn=(1-1/n)^2-a(1-1/n)^2=1/n^2+(a-2)/n+1-a，b(n+1)-bn=(1/(n+1)-1/n)(1/(n+1)+1/n+a-2)=-[1/n(n+1)][1/(n+1)+1/n+a-2]>0

i.e. 1 (n+1)+1 n+a-2)<0

i.e. a<2-1 (n+1)-1 n

2-1 (n+1)-1 n} is a series of increasing numbers (think about why?). The minimum value of 2-1 (n+1)-1 n is the value of n=1, 2-1 (n+1)-1 n=2-1 2-1=1 2

a<1/2

Seek satisfaction.

sn-s(n-1)

n 2) This is the sequence of equations a(n)=

a1n=1）

So: s(n+1

sn=(a(n+1)) 2 The same is true. sn

s(n-1)=(a（n）)^2

Get. a(n+1)+a（n）==a(n+1))^2(an)^2

By Li Hui in an>0, then Beihu is inevitable, a(n+1)>0 so. 1=a(n+1)-a（n

i.e. the sequence a(n

It is an equal difference with the first term A1=1 and the tolerance D=1.

i.e. a(n=a1+(n-1)d

1+(n-1)n

Divide by SNS(n-1)1 s(n-1)-1 sn=21 sn-1 s(n-1)=-21 sn, d=21 s1=1 a1=11 sn=2n-1sn=1 (2n-1)s(n-1)=1 (2n-3)n>=2,an=sn-s(n-1)=-2 (2n-1)(2n-3)a1=1 does not conform so n=1,an=1n 2,an=-2 [(2n-1)(2n-3)]

n>=2

sn=2an+2

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

Subtraction. an=2an-2a(n-1)

an=2a(n-1)

So it's a proportional series, q=2

a1=s1=2a1+2

a1=-2, so an=-2 n

Summary. Hello, your title doesn't seem to be described completely enough, please add, or send it over, sorry thank you.

If sn+n = n (an+1) and a2 = 3, the general formula of an is obtained.

Hello, your title doesn't seem to be described completely enough, please add, or send it over, sorry thank you.

How to send it.

That should be if you don't have enough permissions.

This is to find the proportional series or the equal check series.

Ah, I don't know.

Is the information on the question all on it?

Yes. I've sent you a private message, see if you can send ** over.

Yeah. So the general formula for an is: an=2n-1 OK.

That's it for the process, I hope it can help you. If you have no other questions, please review this service in the lower right corner, and look forward to your praise [Bixin], thank you! <>

Finally, I wish you a happy life [Niu Fu to].

a1=1,an=sn+1 sn(sn+1+sn) to find the general formula for an.

First, according to the idea, we get the recursive formula: an = sn+1 sn(sn+1+sn) Then, we break down the term sn+1 sn, and get: an = 1 sn - 1 (sn+1) Next, we consider how to represent an as a function of n.

Observing the recursive formula, we find that the denominator of each term is the sum of the gross numerator and denominator of the previous term, so we guess that sn = a n + b (n+1), where a and b are the coefficients to be determined. To solve for a and b, we can use the known condition a1=1, i.e.,

s1 = a 1 + b 2 = 1 solves a=2, b=0, therefore: sn = 2 n substituting sn into the formula of an, we get: an = 1 sn - 1 (sn+1) =n 2 - n+1) 2 = 1 2) Therefore, the general formula for an is :

an = 1/2

sn is the sum of the first n terms of {an}, known as an 0An 2+2an=4sn+3 (1) Find the general formula for {an}.

From an2+2an=4sn+3, we can see that an-12+2an-1=4sn-1+3, (n 2) -obtains:an2-an-12+2an-2an-1=4an, i.e., (an+an-1)(an-an-1)=2(an+an-1) an>0, an+an-1≠0, an-an-1=2(n 2), is a series of equal differences with a1=3 as the first term and d=2 as the tolerance an=2n+1(n n*)

s(n+1)=3sn +2

s(n+1)+1=3[s(n)+1]

It's a proportional series.

The first term is s1+1=2, and the common ratio is 3

s(n)+1=2*3^(n-1)

i.e. s(n)=2*3 (n-1)-1

1）n=1,a1=s1=1

2）n≥2,an=s(n)-s(n-1)=2*3^(n-1)-2*3^(n-2)=4*3^(n-2)

In summary. an= 1 n=1

4*3^(n-2) n≥2

an=sn-s(n-1)(n>=2)

3an-3a(n-1)(n>=2)

an/a(n-1)=3/2.

When n=1, a1=3a1+2<=>a1=-1

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

3/2)^(n-1).(n>=2).

Verify that a1=-1 also conforms to this general term.

Therefore, the general term of an is accepted by the formula an=-(3 2) (n-1).

Note: Be sure to verify the A1 term, because some of them do not necessarily satisfy the general term formula that the first term starts from the second term.