大师作文网已知等差数列{an}的首项a1=1,公差d>0,数列{bn}是等比数列,且a2=b2,a5=b3,a14=b4
来源：学生作业帮 编辑：大师作文网作业帮 分类：数学作业 时间：2024/11/11 05:08:06
已知等差数列{an}的首项a1=1,公差d>0,数列{bn}是等比数列,且a2=b2,a5=b3,a14=b4
(1)求数列{an}{bn}通项公式
（2）求数列{anbn}前n项和Tn
(1)求数列{an}{bn}通项公式
（2）求数列{anbn}前n项和Tn
(1)
an=a1+(n-1)d,a1=1
bn=b1q^(n-1)
a2=b2
1+d = b1.q (1)
a5=b3
1+4d = b1q^2 (2)
a14=b4
1+13d = b1q^3 (3)
(3)/(2) = (2)/(1)
(1+13d)/(1+4d) = (1+4d)/(1+d)
1+14d+13d^2= 1+8d+16d^2
d^2-2d=0
d=2
(2)/(1)
q= 3
from (1)
1+2= 3b1
b1=1
an = 1+2(n-1) = 2n-1
bn = 3^(n-1)
(2)
let
S = 1.3^0+2.3^1+...+n.3^(n-1) (1)
3S = 1.3^1+2.3^2+...+n.3^n (2)
(2)-(1)
2S = n.3^n -[1+3+...+3^(n-1) ]
= n.3^n - (3^n -1)/2
cn = an.bn
= (2n-1).3^(n-1)
= 2[n.3^(n-1) ] - 3^(n-1)
Tn = c1+c2+...+cn
=2S - (3^n -1)/2
=n.3^n - (3^n -1)
= 1 + (n-1).3^n
an=a1+(n-1)d,a1=1
bn=b1q^(n-1)
a2=b2
1+d = b1.q (1)
a5=b3
1+4d = b1q^2 (2)
a14=b4
1+13d = b1q^3 (3)
(3)/(2) = (2)/(1)
(1+13d)/(1+4d) = (1+4d)/(1+d)
1+14d+13d^2= 1+8d+16d^2
d^2-2d=0
d=2
(2)/(1)
q= 3
from (1)
1+2= 3b1
b1=1
an = 1+2(n-1) = 2n-1
bn = 3^(n-1)
(2)
let
S = 1.3^0+2.3^1+...+n.3^(n-1) (1)
3S = 1.3^1+2.3^2+...+n.3^n (2)
(2)-(1)
2S = n.3^n -[1+3+...+3^(n-1) ]
= n.3^n - (3^n -1)/2
cn = an.bn
= (2n-1).3^(n-1)
= 2[n.3^(n-1) ] - 3^(n-1)
Tn = c1+c2+...+cn
=2S - (3^n -1)/2
=n.3^n - (3^n -1)
= 1 + (n-1).3^n
已知等差数列{an}的首项a1=1,公差d>0,数列{bn}是等比数列,且a2=b2,a5=b3,a14=b4
已知等差数列{an}的首项a1=1,公差d>0,{bn}为等比数列,且a2=b2,a5=b3,a14=b4,求{an},
已知等差数列{an}的首项a1=1,公差d＞0,等比数列{bn},满足b2=a2,b3=a5,b4=a14．
已知等差数列{an}的首项为a1=1,公差d不为0,等比数列{bn}满足b2=a2,b3=a5,b4=a14
已知等差数列的首项A1=1,公差d>0,A2,A5,A14分别是等比数列的B2,B3,B4
已知等差数列An中A1=1,公差D＞0,且A2,A5,A14分别是等比数列Bn的第二项,三,四项
已知等差数列{an}中 a1=1 公差d>0 且a2 a5 a14 成等比数列 求数列{an}的通项公式 设数列{an}
已知等差数列{an}中,a1=1,公差d>0.且a2,a5,a14分别是等比数列{bn}的第二项,第三项,第四项；
已知等差数列{an}中,a1=1,公差d>0.且a2,a5,a14分别是等比数列{bn}的第二项,第三项,第四项.
已知{an}是公差不为0的等差数列,{bn}是等比数列,其中a1=3,b1=1,a2=b2,3a5=b3,且存
已知等差数列{an}的公差d不等于0,数列{bn}是等比数列,a1=b1=1,a2=b2,a4=b4
已知等差数列｛an｝的公差d不等于0,数列｛bn｝是等比数列,a1=b1=1,a2=b2,a4=b4