题目内容

在正项数列{an}中,令Sn=
n
i=1
1
ai
+
ai+1

(Ⅰ)若{an}是首项为25,公差为2的等差数列,求S100
(Ⅱ)若Sn=
nP
a1
+
an+1
(P为正常数)对正整数n恒成立,求证{an}为等差数列;
(Ⅲ)给定正整数k,正实数M,对于满足a12+ak+12≤M的所有等差数列{an},求T=ak+1+ak+2+…a2k+1的最大值.
(Ⅰ)由题意,利用等差数列的公差为2,得到
1
ai
+
ai+1
=
ai+1
-
ai
2

所以S100=
a101
-
a1
2
=
25+2×100
-
25
2
=5

(Ⅱ)证:令n=1得到
p
a1
+
a2
=
1
a1
+
a2
,则p=1.
由于Sn=
n
i=1
1
ai
+
ai+1
=Sn=
nP
a1
+
an+1
(1),
Sn+1=
n+1
i=1
1
ai
+
ai+1
=
(n+1)P
a1
+
an+2
(2),
(2)-(1),将p=1代入整理得
(n+1)
a1
+
an+2
-
n
a1
+
an+1
=
1
an+1
+
an+2

化简得(n+1)an+1-nan+2=a1(3)
(n+2)an+2-(n+1)an+3=a1(4),
(4)-(3)得an+1+an+3=2an+2对任意的n≥1都成立.
在(3)中令n=1得到,a1+a3=2a2,从而{an}为等差数列.
(Ⅲ)记t=ak+1,公差为d,
则T=ak+1+ak+2+…a2k+1=(k+1)t+
k(k+1)
2
d
,则
T
k+1
=t+
kd
2
,M≥a12+ak+12=t2+(t-kd)2=
4
10
(t+
kd
2
)2+
1
10
(4t-3kd)2
4
10
(t+
kd
2
)2=
2
5
(
T
k+1
)2

T≤
(k+1)
10M
2

当且仅当
4t=3kd
M=
2
5
(t+
kd
2
)2
,即
ak+1=t=3
M
10
d=
4
k
M
10
时等号成立.
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