设数列{an}满足an≠0,a1=1,an=(1-2n)anan-1+an-1(n≥2),数列{an}的前n项和为Sn. (1)求数列{an}的通项公式; (2)求证:当n≥2时,
(3)试探究:当n≥2时,是否有
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(1)∵an≠0
∴anan-1≠0(n≥2)
∴
=an anan-1
+(1-2n)anan-1 anan-1
,an-1 anan-1
即
=(1-2n)+1 an-1
即有1 an
-1 an
=2n-1,1 an-1
∴
=1 an
+(1 a1
-1 a2
)+(1 a1
-1 a3
)+…+(1 a2
-1 an
)=1+3+5+7+…+(2n-1)=1 an-1
=n2(n≥2)n(1+2n-1) 2
又
=1也适合上式,1 a1
∴an=
.1 n2
(2)证明:∵an=1 n2
∴Sn=a1+a2+…+an=1+
+1 22
+…+1 32 1 n2
∵当n≥2时,
<1 n2
=1 (n-1)n
-1 n-1 1 n
∴1+
+1 22
+…+1 32
<1+[(1-1 n2
)+(1 2
-1 2
)+…+(1 3
-1 n-1
)]=2-1 n+1
<2.1 n+1
又∵
>1 n2
=1 n(n+1)
-1 n 1 n+1
∴Sn>(1-
)+(1 2
-1 2
)+…+(1 3
-1 n
)=1-1 n+1
=1 n+1 n n+1
∴当n≥2时,
<Sn<2.n n+1
(3)∵
=1 n2
<4 4n2
=2(4 (2n-1)(2n+1)
-1 2n-1
)1 2n+1
∴1+
+1 22
+…+1 32
<1+2[(1 n2
-1 3
)+(1 5
-1 5
)+…+(1 7
-1 2n-1
)]1 2n+1
=
-5 3
<2 2n+1 5 3
当n≥2时,要Sn>
只需6n (n+1)(2n+1)
>n n+1 6n (n+1)(2n+1)
即需2n+1>6,显然这在n≥3时成立
而S2=1+
=1 4
,当n≥2时5 4
=6n (n+1)(2n+1)
=6×2 (2+1)(4+1)
显然4 5
>5 4 4 5
即当n≥2时Sn>
也成立6n (n+1)(2n+1)
综上所述:当n≥2时,有
<Sn<6n (n+1)(2n+1)
.5 3