已知数列{an}的前n项和Sn=-12n2+kn(其中k∈N+),且Sn的最大值为8.(1)确定常数k,求an;(2)求数列{9-2an2n}的前n项和Tn.-数学

题目简介

已知数列{an}的前n项和Sn=-12n2+kn(其中k∈N+),且Sn的最大值为8.(1)确定常数k,求an;(2)求数列{9-2an2n}的前n项和Tn.-数学

题目详情

已知数列{an}的前n项和Sn=-
1
2
n2+kn(其中k∈N+),且Sn的最大值为8.
(1)确定常数k,求an
(2)求数列{
9-2an
2n
}
的前n项和Tn
题型:解答题难度:中档来源:江西

答案

(1)当n=k时,Sn=-class="stub"1
2
n2+kn
取得最大值
8=Sk=-class="stub"1
2
k2+k2
=class="stub"1
2
k
2
=8
∴k=4,Sn=-class="stub"1
2
n2+4n
从而an=sn-sn-1=-class="stub"1
2
n2+4n
-[-class="stub"1
2
(n-1)2+4(n-1)]=class="stub"9
2
-n

又∵a1=S1=class="stub"7
2
适合上式
an=class="stub"9
2
-n

(2)∵bn=
9-2an
2n
=class="stub"n
2n-1

Tn=1+class="stub"2
2
+class="stub"3
22
+…+class="stub"n-1
2n-2
+class="stub"n
2n-1

class="stub"1
2
Tn
=class="stub"1
2
+ class="stub"2
22
+…+class="stub"n-1
2n-1
+class="stub"n
2n

两式向减可得,class="stub"1
2
Tn=1+class="stub"1
2
+class="stub"1
22
+…+class="stub"1
2n-1
-class="stub"n
2n

=
1-class="stub"1
2n
1-class="stub"1
2
-class="stub"n
2n
=2-class="stub"1
2n-1
-class="stub"n
2n-1

Tn=4-class="stub"n+2
2n-1

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