已知数列{an}的前n项和为sn,且sn=2n^2+n,n是正整数,又an=4log(2)bn+3
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/08/17 11:01:01
已知数列{an}的前n项和为sn,且sn=2n^2+n,n是正整数,又an=4log(2)bn+3
(1)求an,bn
(2)数列{an*bn}的前n项和Tn
我第二小题不会做,求思路
(1)求an,bn
(2)数列{an*bn}的前n项和Tn
我第二小题不会做,求思路
![已知数列{an}的前n项和为sn,且sn=2n^2+n,n是正整数,又an=4log(2)bn+3](/uploads/image/z/8706713-41-3.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BAsn%2C%E4%B8%94sn%3D2n%5E2%2Bn%2Cn%E6%98%AF%E6%AD%A3%E6%95%B4%E6%95%B0%2C%E5%8F%88an%3D4log%EF%BC%882%EF%BC%89bn%2B3)
(1)
Sn= 2n^2+n (1)
S(n-1) = 2(n-1)^2 + (n-1) (2)
(1) -(2)
an= 2n^2+n - 2(n-1)^2 - (n-1)
= 4n-1
an=4log(2)bn+3
4n-1 = 4log(2)bn+3
log(2)bn = n-1
bn = 2^(n-1)
(2)
anbn = (4n-1)(2^(n-1))
= 4[n2^(n-1)] - 2^(n-1)
consider
[x^(n+1)-1]/(x-1) = 1+x+x^2+...+x^n
[(x^(n+1)-1)/(x-1)]' = 1+2x+..+nx^(n-1)
1+2x+..+nx^(n-1) = [ (x-1)(n+1)x^n) -(x^(n+1)-1) ]/(x-1)^2
= [nx^(n+1)-(n+1)x^n+1]/(x-1)^2
put x=2
1+2(2)+3(2)^2+..+n(2)^(n-1)
=n2^(n+1)- (n+1)2^n +1
summation anbn
=summation {4[n2^(n-1)] - 2^(n-1)}
= 4(n2^(n+1)- (n+1)2^n +1) - (2^n-1)
=(4n).2^(n+1) - (4n+5).2^n +5
Sn= 2n^2+n (1)
S(n-1) = 2(n-1)^2 + (n-1) (2)
(1) -(2)
an= 2n^2+n - 2(n-1)^2 - (n-1)
= 4n-1
an=4log(2)bn+3
4n-1 = 4log(2)bn+3
log(2)bn = n-1
bn = 2^(n-1)
(2)
anbn = (4n-1)(2^(n-1))
= 4[n2^(n-1)] - 2^(n-1)
consider
[x^(n+1)-1]/(x-1) = 1+x+x^2+...+x^n
[(x^(n+1)-1)/(x-1)]' = 1+2x+..+nx^(n-1)
1+2x+..+nx^(n-1) = [ (x-1)(n+1)x^n) -(x^(n+1)-1) ]/(x-1)^2
= [nx^(n+1)-(n+1)x^n+1]/(x-1)^2
put x=2
1+2(2)+3(2)^2+..+n(2)^(n-1)
=n2^(n+1)- (n+1)2^n +1
summation anbn
=summation {4[n2^(n-1)] - 2^(n-1)}
= 4(n2^(n+1)- (n+1)2^n +1) - (2^n-1)
=(4n).2^(n+1) - (4n+5).2^n +5
已知数列{an}的前n项和为sn,且sn=2n^2+n,n是正整数,又an=4log(2)bn+3
已知数列{an}的前n项和Sn=-an-(1/2)^(n-1)+2(n为正整数).令bn=2^n*an,求证数列{bn}
已知数列{An},Sn是其前n项和,且满足3An=2Sn+n,n为正整数,求证数列{An+1/2}为等比数列
已知数列an前n项和为Sn,且满足4(n+1)(Sn+1)=(n+2)^2an(n属于正整数) 求an
已知数列{an}的前n和为Sn,且Sn=2an+n^2-3n-2 n为正整数求证:1数列是等比数列2设bn=an*cos
已知数列{an}的前n项和为Sn,满足2Sn+3=3an(n是正整数),{bn}是等差数列,且b2=a1,b4=a1+4
已知数列{an}的前n项和为Sn,且Sn=2n^2+n,n∈N*,数列{bn}满足an=4log2(bn),n∈N*
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
设等差数列{an}的前n项和为Sn,且Sn=((an+1)/2)平方(n属于正整数),若bn=(-1)^nSn,求数列{
已知数列{an}前n项和为sn,且sn=2n^2+n数列{bn}满足an=4log2(bn)+3,n∈N*
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*.
已知数列{an}的前n项和为Sn,且满足Sn=2an-1,n为正整数,求数列{an}的通项公式an