已知数列{an}满足:a1=1,a2=a(a>0),数列{bn}满足:bn=anan+2(n∈N*)
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已知数列{an}满足:a1=1,a2=a(a>0),数列{bn}满足:bn=anan+2(n∈N*)
(1)若数列{an}是等差数列,且b3=45,求a的值及数列{an}通项公式;
(2)若数列{an}的等比数列,求数列{bn}的前n项和Sn.
(1)若数列{an}是等差数列,且b3=45,求a的值及数列{an}通项公式;
(2)若数列{an}的等比数列,求数列{bn}的前n项和Sn.
![已知数列{an}满足:a1=1,a2=a(a>0),数列{bn}满足:bn=anan+2(n∈N*)](/uploads/image/z/7652003-59-3.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3%EF%BC%9Aa1%3D1%EF%BC%8Ca2%3Da%EF%BC%88a%EF%BC%9E0%EF%BC%89%EF%BC%8C%E6%95%B0%E5%88%97%7Bbn%7D%E6%BB%A1%E8%B6%B3%EF%BC%9Abn%3Danan%2B2%EF%BC%88n%E2%88%88N%2A%EF%BC%89)
(1)∵{an}是等差数列,a1=1,a2=a(a>0),∴an=1+(n-1)(a-1).
又b3=45,∴a3a5=45,即(2a-1)(4a-3)=45,
解得a=2或a=-
7
4(舍去),…(5分)
∴an=2n-1.…(7分)
(2))∵{an}是等比数列,a1=1,a2=a(a>0),∴an=an-1,则bn=anan+2=a2n.
∴当a=1时,Sn=n;
当a≠1时,Sn=
a2(1−a2n)
1−a2.…(14分)
又b3=45,∴a3a5=45,即(2a-1)(4a-3)=45,
解得a=2或a=-
7
4(舍去),…(5分)
∴an=2n-1.…(7分)
(2))∵{an}是等比数列,a1=1,a2=a(a>0),∴an=an-1,则bn=anan+2=a2n.
∴当a=1时,Sn=n;
当a≠1时,Sn=
a2(1−a2n)
1−a2.…(14分)
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