已知数列{an}满足a1=1,a2=3,an+2=3an+1-2an(n∈N*). (Ⅰ)证明:数列{an+1-an}是等比数列; (Ⅱ)求数列{an}的通项公式.
已知数列{an}满足a1=1,a2=3,an+2=3an+1-2an(n∈N*).
(Ⅰ)证明:数列{an+1-an}是等比数列;
(Ⅱ)求数列{an}的通项公式.
(Ⅰ)证明:数列{an+1-an}是等比数列;
(Ⅱ)求数列{an}的通项公式.
数学人气:539 ℃时间:2019-10-17 05:36:20
优质解答
证明:(Ⅰ)∵an+2=3an+1-2an,∴an+2-an+1=2(an+1-an),∴an+2−an+1an+1−an=2(n∈N*).∵a1=1,a2=3,∴数列{an+1-an}是以a2-a1=2为首项,2为公比的等比数列.(Ⅱ)由(Ⅰ)得an+1-an=2n(n∈N*),∴an=(...
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