若数列{An}满足前n项之和Sn=2An-4(n∈N*),Bn+1=An+2Bn,且B1=2
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/05/12 13:56:19
若数列{An}满足前n项之和Sn=2An-4(n∈N*),Bn+1=An+2Bn,且B1=2
(1)求数列{An}的通项公式An
(2)求证数列{(Bn)/2^n}是等差数列,并求Bn
(3)求数列{Bn}的前N项个Tn
麻烦写详细过程
(1)求数列{An}的通项公式An
(2)求证数列{(Bn)/2^n}是等差数列,并求Bn
(3)求数列{Bn}的前N项个Tn
麻烦写详细过程
(1)Sn=2An-4……(1),n=1代入,得:A1=4,
Sn+1=(2An+1)-4……(2),
(2)-(1),得:An+1=2An+1-2An,An+1=2An,
数列{An}是首项为4,公比为2的等比数列,An=2^(n+1)
(2)Bn+1=An+2Bn=2Bn+2^(n+1)
Bn+1/2^(n+1)=(Bn/2^n)+1
即(Bn)/2^n是公差为1的等差数列,首项为B1/2^1=2/2=1,通项 (Bn)/2^n=n
(3)(Bn)/2^n=n,Bn=n*2^n
Tn=1*2^1+2*2^2+……+n*2^n
2Tn=1*2^2+2*2^3+……+(n-1)*2^n+n*2^(n+1)
相减得,-Tn=2^1+2^2+2^3+……+2^n-n*2^(n+1)
=2^(n+1)-2-n*2^(n+1)
=(1-n)*2^(n+1)-2
Tn=(n-1)*2^(n+1)+2
Sn+1=(2An+1)-4……(2),
(2)-(1),得:An+1=2An+1-2An,An+1=2An,
数列{An}是首项为4,公比为2的等比数列,An=2^(n+1)
(2)Bn+1=An+2Bn=2Bn+2^(n+1)
Bn+1/2^(n+1)=(Bn/2^n)+1
即(Bn)/2^n是公差为1的等差数列,首项为B1/2^1=2/2=1,通项 (Bn)/2^n=n
(3)(Bn)/2^n=n,Bn=n*2^n
Tn=1*2^1+2*2^2+……+n*2^n
2Tn=1*2^2+2*2^3+……+(n-1)*2^n+n*2^(n+1)
相减得,-Tn=2^1+2^2+2^3+……+2^n-n*2^(n+1)
=2^(n+1)-2-n*2^(n+1)
=(1-n)*2^(n+1)-2
Tn=(n-1)*2^(n+1)+2
3.设数列{an}的前n项和Sn=2an-4(n∈N+),数列{bn}满足:bn+1=an+2bn,且b1=2.求{bn
若数列{an]满足前n项和Sn=2an-4,bn+1=an+2bn,且b1=2,求:bn;{bn}的前n项和Tn
求一道高中数列题若数列{an}满足前n项之和sn=2an-4(n是正整数),b(n+1)=an+2bn,且b1=2,求1
数列an的前n项和为Sn,Sn=4an-3,①证明an是等比数列②数列bn满足b1=2,bn+1=an+bn.求数列bn
数列{an}的前n项的和Sn=2an-1(n∈N*),数列{bn}满足:b1=3,bn+1=an+bn(n∈N*).
an=2*3^n-1 若数列bn满足bn=an+(-1)^n*ln(an),求数列bn前n项和Sn
数列an的前n项和为Sn,Sn=2an-1,数列bn满足b1=2,bn+1=an+bn.求数列bn的前n项和Tn
已知数列an bn其中a1=1/2数列an的前n项和Sn=n^2an(n≥1) 数列bn满足b1=2 bn+1=2bn
设数列an前n项和为Sn,且an+Sn=1,求an的通项公式 若数列bn满足b1=1且bn+1=bn+an,求数列bn通
数列{an}的前n项和Sn=2an-1(n≥1),数列{bn}满足b1=3,b(n+1)=an+bn,求数列{bn}的前
已知等比数列{An}的前n项之和Sn=2^n+p 数列{Bn}满足Bn=log2An,求和:Tn=(b1)^2-(b2)
已知数列{an}的前n项和Sn=n2(n∈N*),数列{bn}为等比数列,且满足b1=a1,2b3=b4