Question

已知数列an的前n项和为sn且sn=1/2(3^n-1)

已知数列an的前n项和为sn且sn=1/2(3^n-1),等差数列bn中,bn>0(n∈N*),且b1+b2+b3=15,又a1+b1, a2+b2, a3+b3,成等比数列

.(1)求数列an , bn 的·通项公式.

(2)求数列an+bn的前n项和Tn

解题过程！！11！！！！！！！

Answer 1

(1)
Sn=(1/2)(3^n-1)

n=1, a1= 1
an = Sn - S(n-1)
= 3^(n-1)
let bn = b1+(n-1)d

b1+b2+b3=15

3b1+3d =15
b1+d =5 (1)

a1+b1, a2+b2, a3+b3,成等比数列
(a1+b1).(a3+b3)=(a2+b2)^2
(1+b1)(9+b1+2d)= (3+b1+d)^2
(1+5-d)(9+5+d)= (3+5)^2
(6-d)(14+d)=64
d^2+8d-20=0
(d+10)(d-2)=0
d=2

from (1) =>b1=3
bn = 3+2(n-1) = 2n+1
(2)
cn = an+bn
= 3^(n-1) + 2n-1
Tn = c1+c2+...+cn
= (1/2)(3^n -1 ) + n^2

Answer 2

（1）
an=sn-s(n-1)=3^n+k-3^(n-1)-k=3^n-3^n/3=2*3^n/3
a1=s1
2=3+k
k=-1
(2)
a(n+1)/2=[2*3^(n+1)/3]/2=3^n
(4+k)^anbn=3^(anbn)
anbn=n
(2*3^n)/3*bn=n
bn=3n/(2*3^n)=3/2*(n/3^n)