已知数列{an}的前n项和为Sn,且Sn=2n2+n,n∈N*,数列{bn}满足an=4log2bn+3,n∈N*.
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已知数列{an}的前n项和为Sn,且Sn=2n2+n,n∈N*,数列{bn}满足an=4log2bn+3,n∈N*.
(1)求an,bn;
(2)求数列{an•bn}的前n项和Tn.
(1)求an,bn;
(2)求数列{an•bn}的前n项和Tn.
(Ⅰ)由Sn=2n2+n可得,当n=1时,a1=s1=3
当n≥2时,an=sn-sn-1=2n2+n-2(n-1)2-(n-1)=4n-1
而n=1,a1=4-1=3适合上式,
故an=4n-1,
又∵an=4log2bn+3=4n-1
∴bn=2n−1
(Ⅱ)由(Ⅰ)知,anbn=(4n−1)•2n−1
Tn=3×20+7×2+…+(4n−1)•2n−1
2Tn=3×2+7×22+…+(4n-5)•2n-1+(4n-1)•2n
∴Tn=(4n−1)•2n−[3+4(2+22+…+2n−1)]
=(4n-1)•2n−[3+4•
2(1−2n−1)
1−2]
=(4n-1)•2n-[3+4(2n-2)]=(4n-5)•2n+5
当n≥2时,an=sn-sn-1=2n2+n-2(n-1)2-(n-1)=4n-1
而n=1,a1=4-1=3适合上式,
故an=4n-1,
又∵an=4log2bn+3=4n-1
∴bn=2n−1
(Ⅱ)由(Ⅰ)知,anbn=(4n−1)•2n−1
Tn=3×20+7×2+…+(4n−1)•2n−1
2Tn=3×2+7×22+…+(4n-5)•2n-1+(4n-1)•2n
∴Tn=(4n−1)•2n−[3+4(2+22+…+2n−1)]
=(4n-1)•2n−[3+4•
2(1−2n−1)
1−2]
=(4n-1)•2n-[3+4(2n-2)]=(4n-5)•2n+5
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