Question

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

Answer 1

(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

Answer 2

（1）
1+d=b1q;
1+4d=b1q²
1+13d=b1q³
三个方程，解三个未知量；
d=0(舍去)
d=2
q=3
b1=1
an=2n-1;
bn=3^(n-1)
（2）
Tn=（2n-1）3^(n-1)
Sn=1*1+3*3+5*9+...+（2n-1）3^(n-1)
3Sn= 1*3+3*9+..... (2n-3)3^(n-1)+(2n-1)3^n
Sn-3Sn=1+2*3+2*9+2*27+....2*3^(n-1)-(2n-1)3^n
-2Sn=3^n-2-1-2n*3^n+3^n=2*3^n-3-2n*3^n
Sn=n*3^n+3/2-3^n
有问题请追问~