(1)∵数列{an}中,a1=1,an+1an-1=anan-1+an2(n∈N+,n≥2),
且
=kn+1,an+1 an
∴
=an+1 an
+1=a n
an-1
+1+1=…=an-1 an-2
+n-1,a2 a1
∵
=k+1,a2 a1
∴
=n+k=kn+1,an+1 an
∴(n-1)k=n-1,
∴k=1.
(2)∵
=kn+1,k=1,an+1 an
∴
=n-1+1=n,an an-1
∴an=n?an-1=n(n-1)?an-2=…=n!,
∵g(x)=
=
anxn-1
(n-1)!
=n?xn-1,n!?xn-1
(n-1)!
∴g(1)=n,g (2)=n? 2n-1,g(3)=n?3n-1,…,g(n)=n?nn-1,
∵f(x)是数列{g(x)}的前n项和,
∴f(x)=g(1)+g(2)+g(3)+…+g(n)
=n+n?2n-1+n?3n-1+…+n?nn-1
=n(1+2n-1+3n-1+…+nn-1).
证明:(3)∵f(2)=n(1+2n-1),
g(3)=3 n
?n?3n-1=3n,3 n
∴不等式f(2)<
g(3)对n∈N+恒成立等价于n(1+2n-1)<33 n