数列{a}满足：a0=1a_0=1a0=1, an+1=∑i1+i2+...+ik=n∏j=1n(Aij),∀n>=0a_{n+1}=\sum_{i_1+i_2+...+i_k=n}^{}{\prod \limits_{j=1}^{n} \left( A_{i_j}\right) },{\forall}n>=0an+1=∑i1+i2+...+ik=nj=1∏n(Aij),∀n>=0 求ana_nan 的通项
数列{a}满足：a0=1a_0=1a0=1,
an+1=∑i1+i2+...+ik=n∏j=1n(Aij),∀n>=0a_{n+1}=\sum_{i_1+i_2+...+i_k=n}^{}{\prod \limits_{j=1}^{n} \left( A_{i_j}\right) },{\forall}n>=0an+1=∑i1+i2+...+ik=nj=1∏n(Aij),∀n>=0
求ana_nan 的通项
