(平)已知数列{an}中,a1=1,an+1an-1=anan-1+an2(n∈N+,n≥2),且an+1an=kn+1.(1)求实数k的值;(2

2025-05-04 01:05:57
推荐回答(1个)
回答(1):

(1)∵数列{an}中,a1=1,an+1an-1=anan-1+an2(n∈N+,n≥2)

an+1
an
=kn+1,
an+1
an
=
a n
an-1
+1
=
an-1
an-2
+1+1
=…=
a2
a1
+n-1

a2
a1
=k+1

an+1
an
=n+k=kn+1

∴(n-1)k=n-1,
∴k=1.
(2)∵
an+1
an
=kn+1
,k=1,
an
an-1
=n-1+1=n

∴an=n?an-1=n(n-1)?an-2=…=n!,
g(x)=
anxn-1
(n-1)!
=
n!?xn-1
(n-1)!
=n?xn-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),
3
n
g(3)=
3
n
?n?3n-1=3n

∴不等式f(2)<
3
n
g(3)
对n∈N+恒成立等价于n(1+2n-1)<3