China Mathematical Competition

(1) When $n=1$, we have $a_1^2=a_1^3$. Since $a_1\ne 0$, we get $a_1=1$. When $n=2$, we have $(1+a_2{)}^2=1+a_2^3$. Since $a_2\ne 0$, we get $a_2=2$ or $a_2=-1$. When $n=3$, we have $(1+a_2+a_3{)}^2=1+a_2^3+a_3^3$. For $a_2=2$, we get $a_3=3$ or $a_3=-2$; for $a_2=-1$, we get $a_3=1$. In summary, we get three sequences consisting of three terms that satisfy the given condition: $$\{1,2,3\},\{1,2,-2\},\text{ and }\{1,-1,1\}.$$

(2) Let $S_n=a_1+a_2+\cdots +a_n$. Then we have $$\begin{gathered} S_n^2=a_1^3+a_2^3+\cdots +a_n^3(n\in \mathbb{N}), \\ (S_n+a_{n+1}{)}^2=a_1^3+a_2^3+\cdots +a_n^3+a_{n+1}^3. \end{gathered}$$ Finding out the difference of the two expressions above and by $a_{n+1}\ne 0$, we have $2S_n=a_{n+1}^2-a_{n+1}$. When $n=1$, we know from (1) that $a_1=1$. When $n\ge 2$, we have $$2a_n=2(S_n-S_{n-1})=(a_{n+1}^2-a_{n+1})-(a_n^2-a_n).$$ And that is $$(a_{n+1}+a_n)(a_{n+1}-a_n-1)=0.$$ Then we get $a_{n+1}=-a_n$ or $a_{n+1}=a_n+1$. Finally, from $a_1=1$ and $a_{2013}=-2012$, we find the formula of general term for a required sequence as $$a_n=\{\begin{array}{ll}n,& 1\le n\le 2012,\\ 2012(-1{)}^n,& n\ge 2013.\end{array}$$