问题 解答题
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
答案

(1)当n=1时,a1=S1=-

1
2
a21
+
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=-

1
2
a2n
+
1
2
an+21-[-
1
2
a2n-1
+
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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