已知数列{an}的前n项和为Sn,且Sn=2n^2+n,n属于N*,数列{bn}满足an=4log2bn+3,n属于N*
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已知数列{an}的前n项和为Sn,且Sn=2n^2+n,n属于N*,数列{bn}满足an=4log2bn+3,n属于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.
![已知数列{an}的前n项和为Sn,且Sn=2n^2+n,n属于N*,数列{bn}满足an=4log2bn+3,n属于N*](/uploads/image/z/17364333-21-3.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E4%B8%94Sn%3D2n%5E2%2Bn%2Cn%E5%B1%9E%E4%BA%8EN%2A%2C%E6%95%B0%E5%88%97%7Bbn%7D%E6%BB%A1%E8%B6%B3an%3D4log2bn%2B3%2Cn%E5%B1%9E%E4%BA%8EN%2A)
(1) 当n=1时,a1=S1=2+1=3;
当n≥2时,an=Sn-S(n-1)=2n²+n-[2(n-1)²+(n-1)]=4n-1
而当n=1时,a1=4-1=3满足此式,∴an=4n-1 (n∈N+)
an=4log2bn+3=4n-1,∴log2bn=n-1,∴bn=2^(n-1) (n∈N+)
(2)an*bn=(4n-1)×2^(n-1)
∴Tn=3×2^0+7×2^1+11×2^2+…+(4n-1)×2^(n-1) ①
那么 2Tn=3×2^1+7×2^2+……+(4n-5)×2^(n-1)+(4n-1)×2^n ②
①-②,得:-Tn=3×2^0+4×2^1+4×2^2+…+4×2^(n-1)-(4n-1)×2^n
=3+4×[2^1+2^2+…+2^(n-1)]-(4n-1)×2^n
=3+4×2×[1-2^(n-1)]/(1-2)-(4n-1)×2^n
=3+4×2^n-8-(4n-1)×2^n
=-5-(4n-5)×2^n
∴Tn=5+(4n-5)×2^n
当n≥2时,an=Sn-S(n-1)=2n²+n-[2(n-1)²+(n-1)]=4n-1
而当n=1时,a1=4-1=3满足此式,∴an=4n-1 (n∈N+)
an=4log2bn+3=4n-1,∴log2bn=n-1,∴bn=2^(n-1) (n∈N+)
(2)an*bn=(4n-1)×2^(n-1)
∴Tn=3×2^0+7×2^1+11×2^2+…+(4n-1)×2^(n-1) ①
那么 2Tn=3×2^1+7×2^2+……+(4n-5)×2^(n-1)+(4n-1)×2^n ②
①-②,得:-Tn=3×2^0+4×2^1+4×2^2+…+4×2^(n-1)-(4n-1)×2^n
=3+4×[2^1+2^2+…+2^(n-1)]-(4n-1)×2^n
=3+4×2×[1-2^(n-1)]/(1-2)-(4n-1)×2^n
=3+4×2^n-8-(4n-1)×2^n
=-5-(4n-5)×2^n
∴Tn=5+(4n-5)×2^n
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