(1)∵a1=S1=pa1∴a1≠0,且p=1,或a1=0.
若是a1≠0,且p=1,则由a1+a2=S2=2pa2.
∴a1=a2,矛盾.故不可能是:a1≠0,且p=1.由a1=0,得a2≠0.
又a1+a2=S2=2pa2,∴p=
1
2.
(2)∵Sn+1=
1
2(n+1)an+1,Sn=
1
2nan,
∴an+1=
1
2(n+1)an+1−
1
2nan.(n-1)an+1=nan.
当k≥2时,
ak+1
ak=
k
k−1.
∴n≥3时有an=
an
an−1•
an−1
an−2•…•
a3
a2•a2=[n−1/n−2•
n−2
n−3•…•
2
1•a2=(n−1)a2.
∴对一切n∈N*有:an=(n-1)a2.
(3)∵45=S10=10×
1
2×a10=45a2,
∴a2=1. an=n-1(n∈N*).
故f(x)=x+2x2+…+nxn.
∴f(
1
3)=
1
3+
2
32+…+
n
3n].
又3•f(
1
3)=
2
3+
3
32+…+
n
3n−1+1.
∴2•f(