作业帮已知数列an的前n项和为sn,数列bn满足bn=log2(an+1),a1=1且对于任意n大于等于有an=2an-1+1已知数列an的前n项和为sn,数列bn满足bn=log2(an+1),a1=1且对于任意n大于等于2,n属于N*,有an=2an-1 +1（1）求an通项公式
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/04 03:16:03
已知数列an的前n项和为sn,数列bn满足bn=log2(an+1),a1=1且对于任意n大于等于有an=2an-1+1已知数列an的前n项和为sn,数列bn满足bn=log2(an+1),a1=1且对于任意n大于等于2,n属于N*,有an=2an-1 +1（1）求an通项公式
已知数列an的前n项和为sn,数列bn满足bn=log2(an+1),a1=1且对于任意n大于等于有an=2an-1+1
已知数列an的前n项和为sn,数列bn满足bn=log2(an+1),a1=1且对于任意n大于等于2,n属于N*,有an=2an-1 +1（1）求an通项公式（2）若cn=an*bn,求数列cn的前n项和Tn
已知数列an的前n项和为sn,数列bn满足bn=log2(an+1),a1=1且对于任意n大于等于有an=2an-1+1已知数列an的前n项和为sn,数列bn满足bn=log2(an+1),a1=1且对于任意n大于等于2,n属于N*,有an=2an-1 +1（1）求an通项公式
1.
n≥2时,
an=2a(n-1)+1
an+1=2a(n-1)+2=2[a(n-1)+1]
(an+1)/[a(n-1)+1]=2,为定值
a1+1=1+1=2,数列{an +1}是以2为首项,2为公比的等比数列
an +1=2×2^(n-1)=2ⁿ
an=2ⁿ-1
n=1时,a1=2-1=1,同样满足通项公式
数列{an}的通项公式为an=2ⁿ-1
2.
bn=log2(an+1)=log2(2ⁿ)=n
cn=an·bn=n·(2ⁿ-1)=n·2ⁿ-n
Tn=c1+c2+...+cn=(1×2+2×2²+...+n×2ⁿ)-(1+2+...+n)
令Cn=1×2+2×2²+...+n×2ⁿ
则2Cn=1×2²+2×2³+...+(n-1)×2ⁿ+n×2^(n+1)
Cn-2Cn=-Cn=2+2²+...+2ⁿ-n×2^(n+1)
=2×(2ⁿ-1)/(2-1)-n×2^(n+1)
=(1-n)×2^(n+1) -2
Cn=(n-1)×2^(n+1) +2
Tn=Cn-(1+2+...+n)
=(n-1)×2^(n+1) +2 -n(n+1)/2
(1)
an=2a(n-1) +1
an +1=2[a(n-1) +1]
=>{an +1}是等比数列,q=2
an +1=2^(n-1).(a1+1)
=2^n
an=-1+2^n
(2)
bn=log<2>(an+1)
=n
let
S=1.2^1+2.2^2+...+n.2^n...
全部展开
(1)
an=2a(n-1) +1
an +1=2[a(n-1) +1]
=>{an +1}是等比数列,q=2
an +1=2^(n-1).(a1+1)
=2^n
an=-1+2^n
(2)
bn=log<2>(an+1)
=n
let
S=1.2^1+2.2^2+...+n.2^n (1)
2S= 1.2^2+2.2^3+...+n.2^(n+1) (2)
(2)-(1)
S=n.2^(n+1)-(1+2^1+...+2^n)
=n.2^(n+1)-(2^n -1)
= 1+(2n-1).2^n
cn=an.bn
= n(-1+2^n)
=(n.2^n)-n
Tn=c1+c2+...+cn
=S - n(n+1)/2
=(2n-1).2^n - (n^2+n-2)/2
收起