Original title: Let m be a set of partial positive integers, and the sum of the first term a 1 = 1 of the sequence {an} and the first n terms is Sn. It is known that any integer k belongs to m. When n > K, S(n+k)+S(n-k)=2(Sn+Sk) holds.

Let m = {3 3,4} and find the general formula of sequence {an}.

The answers extracted from the Internet: When k ∈ m = {3 3,4} and n > k, Sn+k+Sn -k = 2Sn+2Sk and Sn+1+k+Sn+1-k = 2sn+1+. An+1+k+an+1-k = 2an+1,that is, an+1+k-an+1= an+1-an+60.

Why should we take 8 as the boundary? The main purpose is to make arithmetic series, where n is 3 and 4 respectively, have the same number of arithmetic terms. Otherwise, why bother? Exactly, when n≥8, there are the same number of terms a(n+6).

Firstly, a (n+1+k)-a (n+1) = a (n+1-k) is transformed into a (n+1+k).

Because k ∈ m = {3 3,4}, when k=3, that is, when n > k = 3, a(n+4)+a(n-2)=2a(n+ 1).

When n > 4, a (n+3)+a (n-3) = 2an; When n > 5, a (n+2)+a (n-4) = 2a (n-1); When n > 6, a (n+65438) is deduced: When n≥8, a (n+6), A (n+3), An, A (n-3) and A (n-6) become arithmetic progression.

Therefore, when k=4, that is, n > k = 4, a(n+5)+a(n-3)=2a(n+ 1) and n > 5, a(n+4)+a(n-4)=2an,

When n > 6, a(n+3)+a(n-5)=2a(n- 1), when n > 7, a(n+2)+a(n-6)=2a(n-2), when n > 7, then a(n

When n≥8, a(n+2)-an=an-a(n-2), and when n≥9, a (n+1)-a (n-1) = a (n-1...

This method is not good, a bit like patchwork. There is another solution on the Internet, as follows:

Sn + 3 + Sn -3 = 2(Sn+ S3)，Sn+4+Sn-2 = 2(Sn+ 1+S3)an+4+an-2 = 2an+ 1(n≥4)

Sequence {A3n- 1}, {A3n}, {A3n+ 1} (n ≥ 1) are all arithmetic progression.

Sn- a 1 is the sum of the first terms of the three arithmetic progression, and Sn = an2+bn+c(a, B and C are constants);

S 1 = a 1，Sn + 3 + Sn - 3 =2(Sn+ S3)，Sn+4+Sn-4 = 2(Sn+S4)a+b+c = 1，3b + c = 0，4b + c = 0，a = 1，b = c = 0Sn = N2 an = Sn-Sn- 1(S0 = 0)= N2-(n- 1)2 = 2n-65448