已知数列an的前n项和sn满足sn=p(sn-an)+1/2（p为大于0的常数）且2a1是10a3与3a2的等差中项

(1)
Sn=p(Sn-an)+1/2
n=1,a1 =1/2
n=2
a1+a2 = pa1 +1/2
a2 = p/2
n=3
a1+a2+a3 = p( a1+a2) +1/2
(1+p)/2 +a3 =p(1+p)/2 + 1/2
a3 = p^2/2
10a3+3a2 = 4a1
10(p^2/2) + 3(p/2) = 2
10p^2+3p-4=0
(2p-1)(5p+4) =0
p=1/2
Sn=(1/2)(Sn-an)+1/2
Sn= 1- an
an = Sn -S(n-1)
= a(n-1) - an
an = (1/2) a(n-1)
= (1/2)^(n-1) .a1
= (1/2)^n
(2)
an.bn= 2n+1
bn = (2n+1).2^n
= 2(n.2^n) + 2^n
Tn =b1+b2+...+bn
= 2[∑(i:1->n) i.2^i ] + 2(2^n-1)
let
S = 1.2^1 +2.2^2+...+n.2^n (1)
2S = 1.2^2 +2.2^3+...+n.2^(n+1) (2)
(2)-(1)
S = n.2^(n+1) - ( 2+2^2+...+2^n)
=n.2^(n+1) - 2(2^n-1)
Tn = 2S -2(2^n-1)
= 2n.2^(n+1) - 6(2^n-1)
= 6 + (4n-6).2^n