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

题目

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

答案

(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