已知数列{an}的前n项和为Sn,且Sn=2an?2(n∈N*),(1)求a1,a2的值; &n...
已知数列{an}的前n项和为Sn,且Sn=2an?2(n∈N*),(1)求a1,a2的值;(2)求数列{an}的通项an;(3)设cn=(3n+1)an,求数列{cn}的...
已知数列{an}的前n项和为Sn,且Sn=2an?2(n∈N*),(1)求a1,a2的值; (2)求数列{an}的通项an;(3)设cn=(3n+1)an,求数列{cn}的前n项和Tn.
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(1)由S1=2a1-2=a1得a1=2,
S2=2a2-2=a1+a2,a2=4,
(2)∵Sn=2an-2,Sn-1=2an-1-2,
Sn-Sn-1=an,n≥2,n∈N*,
∴an=2an-2an-1,
∵an≠0,
∴
=2,(n≥2,n∈N*).
即数列{an}是等比数列.
an=2?2n?1=22.
(3)cn=(3n+1)an=(3n+1)2n.
Tn=4×2+7×22+10×23+…+(3n-2)2n-1+(3n+1)2n…①,
2Tn=4×22+7×23+10×24+…+(3n-2)2n+(3n+1)2n+1…②,
①-②得 ?Tn=8+3×(22+23+…+2n)?(3n+1)×2n+1…(10分)
=8+3×
?(3n+1)×2n+1…(11分)
=8-12+3?2n+1-(3n+1)?2n+1…(12分)
=-4+(2-3n)?2n+1,…(13分)
∴Tn=4+(3n?2)?2n+1. …(14分)
S2=2a2-2=a1+a2,a2=4,
(2)∵Sn=2an-2,Sn-1=2an-1-2,
Sn-Sn-1=an,n≥2,n∈N*,
∴an=2an-2an-1,
∵an≠0,
∴
an |
an?1 |
即数列{an}是等比数列.
an=2?2n?1=22.
(3)cn=(3n+1)an=(3n+1)2n.
Tn=4×2+7×22+10×23+…+(3n-2)2n-1+(3n+1)2n…①,
2Tn=4×22+7×23+10×24+…+(3n-2)2n+(3n+1)2n+1…②,
①-②得 ?Tn=8+3×(22+23+…+2n)?(3n+1)×2n+1…(10分)
=8+3×
22(1?2n?1) |
1?2 |
=8-12+3?2n+1-(3n+1)?2n+1…(12分)
=-4+(2-3n)?2n+1,…(13分)
∴Tn=4+(3n?2)?2n+1. …(14分)
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