已知数列数列{an}前n项和Sn=-12n2+kn(其中k∈N*),且Sn的最大值为8.(Ⅰ)确定常数k并求{an}的通项公式;(Ⅱ)若bn=9-2an,求数列{1bnbn+1}前n项和Tn.-数学

题目简介

已知数列数列{an}前n项和Sn=-12n2+kn(其中k∈N*),且Sn的最大值为8.(Ⅰ)确定常数k并求{an}的通项公式;(Ⅱ)若bn=9-2an,求数列{1bnbn+1}前n项和Tn.-数学

题目详情

已知数列数列{an}前n项和Sn=-
1
2
n2+kn
(其中k∈N*),且Sn的最大值为8.
(Ⅰ)确定常数k并求{an}的通项公式;
(Ⅱ)若bn=9-2an,求数列{
1
bnbn+1
}
前n项和Tn
题型:解答题难度:中档来源:不详

答案

(Ⅰ)Sn=-class="stub"1
2
n2+kn
=-class="stub"1
2
(n-k)2+class="stub"1
2
k2

又k∈N*,所以当n=k时Sn取得最大值为class="stub"1
2
k2
=8,解得k=4,
Sn=-class="stub"1
2
n2+4n

当n≥2时,an=Sn-Sn-1=(-class="stub"1
2
n2
+4n)-[-class="stub"1
2
(n-1)2+4(n-1)]=-n+class="stub"9
2

当n=1时,a1=-class="stub"1
2
+4=class="stub"7
2
,适合上式,
综上,an=-n+class="stub"9
2

(Ⅱ)由(Ⅰ)得,bn=9-2an=9-2(-n+class="stub"9
2
)=2n,
所以class="stub"1
bnbn+1
=class="stub"1
2n(2n+2)
=class="stub"1
4
(class="stub"1
n
-class="stub"1
n+1
)

Tn=class="stub"1
b1b2
+class="stub"1
b2b3
+…+class="stub"1
bnbn+1
=class="stub"1
4
(1-class="stub"1
2
+class="stub"1
2
-class="stub"1
3
+…+class="stub"1
n
-class="stub"1
n+1
)
=class="stub"1
4
(1-class="stub"1
n+1
)
=class="stub"n
4(n+1)

所以数列{class="stub"1
bnbn+1
}
前n项和Tn为class="stub"n
4(n+1)

更多内容推荐