设数列{an}的各项均为正数,前n项和为Sn,已知4Sn=
(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
(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. |
(1)∵4Sn=
+2an+1,∴当n≥2时,4Sn-1=a 2n
+2an-1+1.a 2n-1
两式相减得4an=
-a 2n
+2an-2an-1,a 2n-1
∴(an+an-1)(an-an-1-2)=0,
∵an>0,∴an-an-1=2,
又4S1=
+2a1+1,∴a1=1,a 21
∴{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=
=n2,(1+2n-1)n 2
∴Sm=m2,Sk=k2,Sp=p2,
于是
+1 Sm
-1 Sp
=2 Sk
+1 m2
-1 p2
=2 k2 k2(p2+m2)-2m2p2 m2p2k2
=
≥(
)2(p2+m2)-2m2p2m+p 2 m2p2k2
=0,mp×2pm-2m2p2 m2p2k2
∴
+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+
d=n(n-1) 2
,n(a1+an) 2
于是Sm+Sp-2Sk=ma1+
d+pa1+m(m-1) 2
d-[2ka1+k(k-1)d]p(p-1) 2
=(m+p)a1+
d-(2ka1+k2d-kd),m2+p2-m-p 2
将m+p=2k代入得,Sm+Sp-2Sk=
d≥0,(m-p)2 4
∴Sm+Sp≥2Sk,
又SmSp=
=mp(a1+am)(a1+ap) 4
≤mp[
+(am+ap)a1+amap]a 21 4 (
)2[m+p 2
+2a1ak+(a 21
)2]am+ap 2 4
=
=k2(a12+2a1ak+
)a 2k 4
=k2(a1+ak)2 4
,S 2k
∴
+1 Sm
=1 Sp
≥Sm+Sp SmSp
=2Sk S 2k
.2 Sk