设单调递减数列{an}前n项和Sn=-12a2n+12an+21,且a1>0;(1)求{an}的通项公式;(2)若bn=2n-1•an,求{bn}前n项和Tn.-高二数学

题目简介

设单调递减数列{an}前n项和Sn=-12a2n+12an+21,且a1>0;(1)求{an}的通项公式;(2)若bn=2n-1•an,求{bn}前n项和Tn.-高二数学

题目详情

设单调递减数列{an}前n项和Sn=-
1
2
a2n
+
1
2
an+21
,且a1>0;
(1)求{an}的通项公式;
(2)若bn=2n-1an,求{bn}前n项和Tn
题型:解答题难度:中档来源:不详

答案

(1)当n=1时,a1=S1=-class="stub"1
2
a21
+class="stub"1
2
a1+21
,化为
a21
+a1-42=0
,又a1>0,解得a1=6;
当n≥2时,an=Sn-Sn-1=-class="stub"1
2
a2n
+class="stub"1
2
an+21
-[-class="stub"1
2
a2n-1
+class="stub"1
2
an-1+21]
,化为(an+an-1)(an-an-1+1)=0,
∵数列{an}是单调递减数列,∴an+an-1≠0,an-an-1=-1.
∴数列{an}是公差为-1的等差数列,∴an=a1+(n-1)d=6-(n-1)=7-n.
(2)∵bn=2n-1an=(7-n)•2n-1.
∴Tn=6×1+5×21+4×22+…+(8-n)×2n-2+(7-n)×2n-1,
2Tn=6×21+5×22+…+(8-n)×2n-1+(7-n)×2n,
Tn=-6+(21+22+…+2n-1)+(7-n)×2n
=-6+
2(2n-1-1)
2-1
+(7-n)×2n
=-6+2n-2+(7-n)×2n
=(8-n)×2n-8..

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