已知各项均为正数的数列{an},其前n项和为Sn,且满足4Sn=(an+1)2(Ⅰ)求数列{an}的通项公式;(Ⅱ)设
已知各项均为正数的数列{an},其前n项和为Sn,且满足4Sn=(an+1)2(Ⅰ)求数列{an}的通项公式;(Ⅱ)设数列{bn}满足:b1=3,bn+1=abn,记cn...
已知各项均为正数的数列{an},其前n项和为Sn,且满足4Sn=(an+1)2(Ⅰ)求数列{an}的通项公式;(Ⅱ)设数列{bn}满足:b1=3,bn+1=abn,记cn=anbn,求数列{cn}的前n项和Tn.
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(本小题满分13分)
(Ⅰ)∵4Sn=(an+1)2当n≥2时,4Sn?1=(an?1+1)2
两式相减得:(an+an-1)(an-an-1-2)=0
又an>0故an-an-1=2,
∴{an}是以2为公差的等差数列
又a1=1,
∴an=2n-1.(6分)
(Ⅱ)∵bn+1=abn=2bn?1,
∴bn+1-1=2(bn-1)
又b1-1=2≠0,∴{bn-1}是以2为公比的等比数列,
∴bn?1=2n,
∴bn=2n+1,
故cn=anbn=(2n?1)2n+(2n?1)
记An=1×2+3×22+…+(2n?1)2n,①
2An=1×22+3×23+…+(2n-1)?2n+1,②
①-②,得:-An=2+22+23+…+2n-(2n-1)?2n+1
=
?(2n?1)?2n+1,
由错位相减得:
An=(2n?3)2n+1+6,
∴Tn=(2n?3)2n+1+n2+6.(13分)
(Ⅰ)∵4Sn=(an+1)2当n≥2时,4Sn?1=(an?1+1)2
两式相减得:(an+an-1)(an-an-1-2)=0
又an>0故an-an-1=2,
∴{an}是以2为公差的等差数列
又a1=1,
∴an=2n-1.(6分)
(Ⅱ)∵bn+1=abn=2bn?1,
∴bn+1-1=2(bn-1)
又b1-1=2≠0,∴{bn-1}是以2为公比的等比数列,
∴bn?1=2n,
∴bn=2n+1,
故cn=anbn=(2n?1)2n+(2n?1)
记An=1×2+3×22+…+(2n?1)2n,①
2An=1×22+3×23+…+(2n-1)?2n+1,②
①-②,得:-An=2+22+23+…+2n-(2n-1)?2n+1
=
2(1?2n) |
1?2 |
由错位相减得:
An=(2n?3)2n+1+6,
∴Tn=(2n?3)2n+1+n2+6.(13分)
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