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.

来源：学生作业帮助网编辑：作业帮时间：2024/04/30 22:11:06

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.
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.

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) 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

已知数列｛an｝的前n项和为sn,且满足sn＝n 已知数列{an}的前n项和为Sn,且Sn=lgn 求通项公式已知数列an是等差数列,且a1不等于0,Sn为这个数列的前n项和,求limnan/Sn.limSn+Sn-1/Sn+Sn-1 1.已知数列an的前n项和为Sn,且Sn=2^n,求通项an；2.已知数列an的前n项和为Sn,且Sn=n^2+3n,求通项an；已知数列an的前n项和为Sn,且An=3^n+2n,则Sn等于已知数列{an}的前n项和为Sn,且an=n2的n次方,则Sn= 已知{an}的前n项和为Sn,且an+Sn=4求证：数列{an}是等比数列已知数列{an}的前n项和为Sn,an+Sn=2,（n 已知数列{an}的前n项和为Sn,且满足an+2Sn*Sn-1=0,a1=1/2.求证：{1/Sn}是等差数列已知数列an的前n项和为Sn,且Sn=2(an-1),则a2等于已知数列{an}的前n项和为Sn,且Sn=3an+1则a4=? 已知数列[AN]的前N项和为SN且A1＝1SN＝N²AN[N∈N'] 猜想SN的表达式并验证已知数列{an}的前n项和为Sn,且Sn=n-5an-85,求an和Sn 已知数列an前n项的和为Sn 且满足Sn=1-nan n=自然数已知数列前n项和为Sn,且Sn=-2n+3,求an及Sn 已知数列{an}的前n项和为Sn,且Sn=2an-n（n∈N*）,求数列{an}的通项公式. 已知数列an的前n项和为sn,且sn+an=n^2+3n+5/2,证明数列{an-n}是等比数列已知数列{an}的前n项和Sn,且(1-k)Sn=1-kan求an、sn