问题
解答题
已知数列{an}的前n项和为Sn,且满足 an+2Sn•Sn-1=0(n≥2),a1=
(1)求证:{
(2)求an表达式; (3)若bn=2(1-n)an(n≥2),求证:b22+b32+…+bn2<1. |
答案
解(1)∵-an=2SnSn-1,
∴-Sn+Sn-1=2SnSn-1(n≥2)
Sn≠0,∴
-1 Sn
=2,又1 Sn-1
=1 S1
=2,1 a1
∴{
}是以2为首项,公差为2的等差数列.1 Sn
(2)由(1)
=2+(n-1)2=2n,1 Sn
∴Sn=1 2n
当n≥2时,an=Sn-Sn-1=-1 2n(n-1)
n=1时,a1=S1=
,1 2
∴an=
;
(n=1)1 2 -
(n≥2)1 2n(n-1)
(3)由(2)知bn=2(1-n)an=1 n
∴b22+b32+…+bn2=
+1 22
+…+1 32
<1 n2
+1 1×2
+…+1 2×3 1 (n-1)n
=(1-
)+(1 2
-1 2
)+…+(1 3
-1 n-1
)=1-1 n
<1.1 n