已知数列{an}满足a1=1,a2=3,an+2=3an+1-2an(n∈N+)
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已知数列{an}满足a1=1,a2=3,an+2=3an+1-2an(n∈N+)
(1)证明:数列{an+1-an }是等比数列;
(2)求数列{an}的通项公式.
(1)证明:数列{an+1-an }是等比数列;
(2)求数列{an}的通项公式.
(1)证明:∵an+2=3an+1-2an
∴an+2-an+1=2(an+1-an)
又a1=1,a2=3
即
an+2−an+1
an+1−an=2
∴数列{an+1-an}是以2为 首项,2为公比的等比数列
(2)由(1)知an+1-an=2n
∴an=(an-an-1)+(an-1-an-2)+…+(a2-a1)+a1
=2n-1+2n-2+…+2+1
=2n-1
∴an+2-an+1=2(an+1-an)
又a1=1,a2=3
即
an+2−an+1
an+1−an=2
∴数列{an+1-an}是以2为 首项,2为公比的等比数列
(2)由(1)知an+1-an=2n
∴an=(an-an-1)+(an-1-an-2)+…+(a2-a1)+a1
=2n-1+2n-2+…+2+1
=2n-1
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