设数列{an}是公差为d的等差数列,其前n项和为Sn.已知a1=1,d=2,①求当n∈N*时,Sn+64n的最小值;②证明:由①知Sn=n2,当n∈N*时,2s1s3+3s2s4…+n+1SnSn+2

题目简介

设数列{an}是公差为d的等差数列,其前n项和为Sn.已知a1=1,d=2,①求当n∈N*时,Sn+64n的最小值;②证明:由①知Sn=n2,当n∈N*时,2s1s3+3s2s4…+n+1SnSn+2

题目详情

设数列{an}是公差为d的等差数列,其前n项和为Sn.已知a1=1,d=2,
①求当n∈N*时,
Sn+64
n
的最小值;
②证明:由①知Sn=n2,当n∈N*时,
2
s1s3
+
3
s2s4
…+
n+1
SnSn+2
5
16
题型:解答题难度:中档来源:不详

答案

①∵a1=1,d=2,∴Sn=na1+
n(n-1)
2
d
=n2,
class="stub"Sn+64
n
=
n2+64
n
=n+class="stub"64
n
2
n×class="stub"64
n
=16
当且仅当n=class="stub"64
n
即n=8时,上式取等号,
class="stub"Sn+64
n
的最小值是16;
②证明:由①知Sn=n2,当n∈N*时,
class="stub"n+1
SnSn+2
=class="stub"n+1
n2(n+2)2
=class="stub"1
4
[class="stub"1
n2
-class="stub"1
(n+2)2
]

class="stub"2
s1s3
+class="stub"3
s2s4
…+class="stub"n+1
SnSn+2

=class="stub"1
4
[class="stub"1
12
-class="stub"1
32
+class="stub"1
22
-class="stub"1
42
+class="stub"1
32
-class="stub"1
52
+…+class="stub"1
n2
-class="stub"1
(n+2)2
]
=class="stub"1
4
[class="stub"1
12
+class="stub"1
22
-class="stub"1
(n+1)2
-class="stub"1
(n+2)2
]

class="stub"1
(n+1)2
+class="stub"1
(n+2)2
>0

class="stub"2
s1s3
+class="stub"3
s2s4
…+class="stub"n+1
SnSn+2
<class="stub"1
4
(class="stub"1
12
+class="stub"1
22
)
=class="stub"5
16

故命题得证.

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