问题
解答题
设单调递减数列{an}前n项和Sn=-
(1)求{an}的通项公式; (2)若bn=2n-1•an,求{bn}前n项和Tn. |
答案
(1)当n=1时,a1=S1=-1 2
+a 21
a1+21,化为1 2
+a1-42=0,又a1>0,解得a1=6;a 21
当n≥2时,an=Sn-Sn-1=-1 2
+a 2n
an+21-[-1 2 1 2
+a 2n-1
an-1+21],化为(an+an-1)(an-an-1+1)=0,1 2
∵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-1•an=(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+
+(7-n)×2n2(2n-1-1) 2-1
=-6+2n-2+(7-n)×2n
=(8-n)×2n-8..