题目内容
在正项数列{an}中,令Sn=
.
(Ⅰ)若{an}是首项为25,公差为2的等差数列,求S100;
(Ⅱ)若Sn=
(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.
n |
∑i=1 |
1 | ||||
|
(Ⅰ)若{an}是首项为25,公差为2的等差数列,求S100;
(Ⅱ)若Sn=
nP | ||||
|
(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
=
,
所以S100=
=
=5.
(Ⅱ)证:令n=1得到
=
,则p=1.
由于Sn=
=Sn=
(1),
Sn+1=
=
(2),
(2)-(1),将p=1代入整理得
-
=
,
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+
d,则
=t+
,M≥a12+ak+12=t2+(t-kd)2=
(t+
)2+
(4t-3kd)2≥
(t+
)2=
(
)2
则T≤
,
当且仅当
,即
时等号成立.
1 | ||||
|
| ||||
2 |
所以S100=
| ||||
2 |
| ||||
2 |
(Ⅱ)证:令n=1得到
p | ||||
|
1 | ||||
|
由于Sn=
n |
∑i=1 |
1 | ||||
|
nP | ||||
|
Sn+1=
n+1 |
∑i=1 |
1 | ||||
|
(n+1)P | ||||
|
(2)-(1),将p=1代入整理得
(n+1) | ||||
|
n | ||||
|
1 | ||||
|
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 |
T |
k+1 |
kd |
2 |
4 |
10 |
kd |
2 |
1 |
10 |
4 |
10 |
kd |
2 |
2 |
5 |
T |
k+1 |
则T≤
(k+1)
| ||
2 |
当且仅当
|
|
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