设数列{an}的前n项和为Sn,且满足S1=2,Sn+1=3Sn+2(n=1,2,3,…).
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设数列{an}的前n项和为Sn,且满足S1=2,Sn+1=3Sn+2(n=1,2,3,…).
(Ⅰ)证明数列{an}是等比数列并求通项an;
(Ⅱ)求数列{nan}的前n项和Tn.
(Ⅰ)证明数列{an}是等比数列并求通项an;
(Ⅱ)求数列{nan}的前n项和Tn.
证明:(Ⅰ)∵Sn+1=3Sn+2,
∴Sn=3Sn-1+2(n≥2)
两式相减得an+1=3an(n≥2)
∵S1=2,Sn+1=3Sn+2
∴a1+a2=3a1+2即a2=6则
a2
a1=3
∴
an+1
an=3(n≥1)
∴数列{an}是首项为2,公比为3的等比数列
∴an=2×3n-1(n=1,2,3,…).
(Ⅱ)∵Tn=1•a1+2•a2+…+nan=1×2+2×2×31+…+n×2×3n-1,
∴3Tn=1×2×3+2×2×32+…+(n-1)×2×3n-1+n×2×3n,(9分)
∴-2Tn=2(1+3+32+…3n-1)-n×2×3n=2×
3n−1
3−1-n×2×3n=3n(1-2n)-1(11分)
∴Tn=
(2n−1)3n+1
2 (13分)
∴Sn=3Sn-1+2(n≥2)
两式相减得an+1=3an(n≥2)
∵S1=2,Sn+1=3Sn+2
∴a1+a2=3a1+2即a2=6则
a2
a1=3
∴
an+1
an=3(n≥1)
∴数列{an}是首项为2,公比为3的等比数列
∴an=2×3n-1(n=1,2,3,…).
(Ⅱ)∵Tn=1•a1+2•a2+…+nan=1×2+2×2×31+…+n×2×3n-1,
∴3Tn=1×2×3+2×2×32+…+(n-1)×2×3n-1+n×2×3n,(9分)
∴-2Tn=2(1+3+32+…3n-1)-n×2×3n=2×
3n−1
3−1-n×2×3n=3n(1-2n)-1(11分)
∴Tn=
(2n−1)3n+1
2 (13分)
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