已知数列an的前n项和为Sn,a1=1,an+1=1+2sn（n属于N）（1）求a2 a3 a4的值

(1) and (2)
a(n+1)= 1+2Sn
for n>=2
an = Sn -S(n-1)
2an = a(n+1) - an
a(n+1) = 3an
an = 3^(n-1) .a1
= 3^(n-1)
a2=3
a3=9
a4=27
(3)
let
S =1.(1/3)^0 +2.(1/3)^1+...+n.(1/3)^(n-1) (1)
(1/3)S = 1.(1/3)^1 +2.(1/3)^2+...+n.(1/3)^n (2)
(1)-(2)
(2/3)S = (1+1/3+...+(1/3)^(n-1)) -n.(1/3)^n
= (3/2)(1-(1/3)^n) -n.(1/3)^n
S = (9/4)(1-(1/3)^n) -(3/2)n.(1/3)^n
=(9/4) - [(3/2)n +9/4].(1/3)^n
bn=n/an
b1+b2+...+bn =S =(9/4) - [(3/2)n +9/4].(1/3)^n