一、由a_n=S_n-S_(n-1)来确定数列的构造思路例1已知数列{a_n}的首项为1,S_n为数列{a_n}的前n项和,S_(n+1)=qS_n+1,其中q>0,n∈N~*.若2a_2,a_3,a_2+2成等差数列,求数列{a_n}的通项公式.解由S_(n-1)=qS_n+1,可得当n≥2时,S_n=qS_(n-1)+1,则a_(n+1)=S_(n+1)-S_n=qa_n.由S_2=a_1+a_2=qa_1+1,可得a_2=qa_1,所以a_(n-1)=qa_n对所有n≥1都成立.故数列{a_n}是以1为首 First, a_n = S_n-S_ (n-1) to determine the structure of the sequence of ideas Example 1 Known sequence {a_n} of the first item is 1, S_n series {a_n} the first n terms and, S_ (n +1 ) = qS_n + 1, where q> 0, n∈N ~ *. If 2a_2, a_3, a_2 + 2 are arithmetic sequence, find the general formula of {a_n} +1, it can be obtained that when n_n≥2, S_n = qS_ (n-1) +1, then a_ (n + 1) = S_ (n + 1) -S_n = qa_n. From S_2 = a_1 + a_2 = qa_1 + 1 , We can get a_2 = qa_1, so a_ (n-1) = qa_n holds for all n≥1, so the sequence {a_n} is headed by 1