已知当x=5时,二次函数f(x)=ax2+bx+c取得最小值,等差数列{an}的前n项和Sn=f(n),a2=-7.(Ⅰ)求数列{an}的通项公式;(Ⅱ)数列{bn}的前n项和为Tn,且bn=an2n

题目简介

已知当x=5时,二次函数f(x)=ax2+bx+c取得最小值,等差数列{an}的前n项和Sn=f(n),a2=-7.(Ⅰ)求数列{an}的通项公式;(Ⅱ)数列{bn}的前n项和为Tn,且bn=an2n

题目详情

已知当x=5时,二次函数f(x)=ax2+bx+c取得最小值,等差数列{an}的前n项和Sn=f(n),a2=-7.
(Ⅰ)求数列{an}的通项公式;
(Ⅱ)数列{bn}的前n项和为Tn,且bn=
an
2n
,证明Tn≤-
9
2
题型:解答题难度:中档来源:临沂三模

答案

(Ⅰ)当n=1时,a1=S1=a+b+c,
当n≥2时,an=Sn-Sn-1=2an+b-a,
又a1适合上式,得2a+b-a=a+b+c,∴c=0.
由已知a2=4a+b-a=3a+b=-7,-class="stub"b
2a
=5

解方程组
3a+b=-7
-class="stub"b
2a
=5
a=1
b=-10

∴an=2n-11.
(Ⅱ)bn=class="stub"2n-11
2n

Tn=class="stub"-9
2
+class="stub"-7
22
+…+class="stub"2n-11
2n
class="stub"1
2
Tn=…class="stub"-9
22
+…+class="stub"2n-13
2n
+class="stub"2n-11
2n+1

①-②得class="stub"1
2
Tn=-class="stub"9
2
+class="stub"2
22
+…+class="stub"2
2n
-class="stub"2n-11
2n+1

=-class="stub"9
2
+
class="stub"1
2
(1-class="stub"1
2n-1
)
1-class="stub"1
2
-class="stub"2n-11
2n+1
=-class="stub"7
2
-class="stub"1
2n-1
-class="stub"2n-11
2n+1

Tn=-7-class="stub"2n-7
2n

T1=-class="stub"9
2
T2=-class="stub"9
2
-class="stub"7
2
<-class="stub"9
2
T3=-class="stub"9
2
-class="stub"7
2
-class="stub"5
2
<-class="stub"9
2

当n≥4时,class="stub"2n-7
2n
>0
,∴Tn=-7-class="stub"2n-7
2n
<-7<-class="stub"9
2

综上,得Tn≤-class="stub"9
2

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