问题 解答题
设数列{an}的各项均为正数,前n项和为Sn,已知4Sn=
a2n
+2an+1(n∈N*)

(1)证明数列{an}是等差数列,并求其通项公式;
(2)证明:对任意m、k、p∈N*,m+p=2k,都有
1
Sm
+
1
Sp
2
Sk

(3)对于(2)中的命题,对一般的各项均为正数的等差数列还成立吗?如果成立,请证明你的结论,如果不成立,请说明理由.
答案

(1)∵4Sn=

a2n
+2an+1,∴当n≥2时,4Sn-1=
a2n-1
+2an-1+1

两式相减得4an=

a2n
-
a2n-1
+2an-2an-1

∴(an+an-1)(an-an-1-2)=0,

∵an>0,∴an-an-1=2,

4S1=

a21
+2a1+1,∴a1=1,

∴{an}是以a1=1为首项,d=2为公差的等差数列.  

∴an=2n-1;

(2)由(1)知Sn=

(1+2n-1)n
2
=n2

Sm=m2Sk=k2Sp=p2

于是

1
Sm
+
1
Sp
-
2
Sk
=
1
m2
+
1
p2
-
2
k2
=
k2(p2+m2)-2m2p2
m2p2k2

=

(
m+p
2
)
2
(p2+m2)-2m2p2
m2p2k2
mp×2pm-2m2p2
m2p2k2
=0

1
Sm
+
1
Sp
2
Sk

(3)结论成立,证明如下:

设等差数列{an}的首项为a1,公差为d,则Sn=na1+

n(n-1)
2
d=
n(a1+an)
2

于是Sm+Sp-2Sk=ma1+

m(m-1)
2
d+pa1+
p(p-1)
2
d-[2ka1+k(k-1)d]

=(m+p)a1+

m2+p2-m-p
2
d-(2ka1+k2d-kd),

将m+p=2k代入得,Sm+Sp-2Sk=

(m-p)2
4
d≥0,

∴Sm+Sp≥2Sk

SmSp=

mp(a1+am)(a1+ap)
4
=
mp[
a21
+(am+ap)a1+amap]
4
(
m+p
2
)
2
[
a21
+2a1ak+(
am+ap
2
)
2
]
4

=

k2(a12+2a1ak+
a2k
)
4
=
k2(a1+ak)2
4
=
S2k

1
Sm
+
1
Sp
=
Sm+Sp
SmSp
2Sk
S2k
=
2
Sk

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