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例1已知数列{a_n}满足a_1=1,a_(n+1)=2a_n+1(n∈N~*),求数列{a_n}的通项公式.解法1由a_(n+1)=2a_n+1,得a_(n+1)+1=2(a_n+1),所以,数列{a_n+1}是首项为2,公比为2的等比数列,
(N + 1) = 2a_n + 1 (n∈N ~ *), find the general formula of the sequence {a_n}. The solution 1 consists of a_ (n + 1) (N + 1) + 1 = 2 (a_n + 1). Thus, the sequence {a_n + 1} is the geometric sequence whose first term is 2 and whose common power is 2,