(1)令n=1得:
S21-(-1)S1-3×2=0,即
S21+S1-6=0.
∴(S1+3)(S1-2)=0.
∵S1>0,∴S1=2,即a1=2.
(2)由
S2n-(n2+n-3)Sn-3(n2+n)=0得:
(Sn+3)[Sn-(n2+n)]=0.
∵an>0(n∈N*),
∴Sn>0.
∴Sn=n2+n.
∴当n≥2时,an=Sn-Sn-1=(n2+n)-[(n-1)2+(n-1)]=2n,
又∵a1=2=2×1,
∴an=2n(n∈N*).
(3)当k∈N*时,∵k(k+
1
2)=k2+
1
2k>k2+
1
2k-
3
16=(k-
1
4)(k+
3
4),
∴[1
ak(ak+1)=
1
2k(2k+1)=
1/4•
1
k(k+
1
2)<
1
4•
1
(k-
1
4)(k+
3
4)]
=
1
4[
1
k-
1
4-
1
(k+1)-
1
4].
∴
1
a1(a1+1)+
1
a2(a