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【题目】在数列{a_n}中,S_(n+1)=4a_n+2,a_1=1.(1)设 b_n=a_(n+1)-2a_n,求证数列{b_n}是等比数列;(2)设 c_n=a_n/2~n,求证{c_n}是等差数列;(3)求数列{a_n}的通项公式及前 n 项和公式.解:(1)由 S_(n+1)=4a_n+2,得 S_(n+2)=4a_(n+1)+2,两式相减,S_(n+2)-S_(n+1)=4(a_(n+1)-a_n),即a_(n+2)=4(a_(n+1)-a_n),变形得 a_(n+2)-2a_(n+1)=2(a_(n+1)-2a_n),即 b_(n+1)=2b_n.又 a_1=1,S_2=4a_1+2=6,∴a_2=S_2-a_1=6-1=5,b_1=a_2-2a_1=3,∴{b_n}是首项为3,公比为2的
[Title] In the sequence {a_n}, S_(n+1)=4a_n+2, a_1=1. (1) Let b_n=a_(n+1)-2a_n, and the series {b_n} of the series of proofs is a geometric sequence; (2) Let c_n=a_n/2~n, and verify that {c_n} is an arithmetic progression; (3) Find the general term of the sequence {a_n} and the first n terms and formulas. Solution: (1) by S_(n+ 1)=4a_n+2, S_(n+2)=4a_(n+1)+2, the two equations are subtracted, S_(n+2)-S_(n+1)=4(a_(n+1) )-a_n), that is, a_(n+2)=4(a_(n+1)-a_n), a_(n+2)-2a_(n+1)=2(a_(n+1)- 2a_n), ie b_(n+1)=2b_n. Again a_1=1, S_2=4a_1+2=6, ∴a_2=S_2-a_1=6-1=5, b_1=a_2-2a_1=3, ∴{b_n } is the first item of 3 and the common ratio is 2