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*.

题目

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.

答案

(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