数列题求解！！感谢大佬！
如图
an = a1.q^(n-1)
a1+a2+...+an=Sn
a1=1, S6=9S3
bn=b1+(n-1)d
b3+b5=a4
b4+2b6=a5
solution :
(I)
S6=9S3
a1( q^6-1)/(q-1) =9a1.(q^3-1)/(q-1)
q^6-1 =9(q^3-1)
q^6-9q^3 +8=0
(q^3-1)(q^3-8)=0
q^3 =8
q=2
an = a1.q^(n-1) = 2^(n-1)
b3+b5=a4
2b1+6d =8 (1)
b4+2b6=a5
3b1 + 13d =16 (2)
2(2)-3(1)
8d=8
d=1
from (1)
2b1+6=8
b1=1
bn = n
(II)
let
S = 1. (1/2)^1 + 2.(1/2)^2+....+n.(1/2)^n (3)
(1/2)S = 1. (1/2)^2 + 2.(1/2)^3+....+n.(1/2)^(n+1) (4)
(3)-(4)
(1/2)S = [(1/2)^1+(1/2)^2 +...+(1/2)^n ] -n.(1/2)^(n+1)
= ( 1 - (1/2)^n ) -n.(1/2)^(n+1)
S = 2( 1 - (1/2)^n ) -n.(1/2)^n
cn
=bn/an
=n/2^n
Tn
=c1+c2+...+cn
=S
=2( 1 - (1/2)^n ) -n.(1/2)^n
=2 -( n +2) .(1/2)^n
