已知各项均为正数的数列{an}的前n项和为Sn,且Sn,an,1成等差数列.(1)求数列{an}的通项公式;(2)若an2=2-bn,设Cn=bnan求数列{Cn}的前项和Tn.-数学

题目简介

已知各项均为正数的数列{an}的前n项和为Sn,且Sn,an,1成等差数列.(1)求数列{an}的通项公式;(2)若an2=2-bn,设Cn=bnan求数列{Cn}的前项和Tn.-数学

题目详情

已知各项均为正数的数列{an}的前n项和为Sn,且Sn,an,1成等差数列.
(1)求数列{an}的通项公式;
(2)若an2=2-bn,设Cn=
bn
an
求数列{Cn}的前项和Tn
题型:解答题难度:中档来源:不详

答案

(1)由题意2an=Sn+1,an>0
当n=1时2a1=a1+1∴a1=1
n≥2时,sn=2an-1,sn-1=2an-1-1
两式相减an=2an-2an-1(n≥2)
整理得
an
an-1
=2
(n≥2)(4分)
∴数列{an}1为首项,2为公比的等比数列.
∴an=a1•2n-1=1×2n-1=2n-1(5分)
(2)an2=2-bn=22n-2
∴bn=2-2n(6分)
Cn=
bn
an
=class="stub"2-2n
2n-1
=class="stub"4-4n
2n
Tn =class="stub"0
2
+class="stub"-4
22
+class="stub"-8
23
+…+class="stub"8-4n
2n-1
+class="stub"4-4n
2n

class="stub"1
2
Tn=class="stub"0
22
+class="stub"-4
23
+…+class="stub"8-4n
2n
+class="stub"4-4n
2n+1

①-②class="stub"1
2
Tn=-4(class="stub"1
22
+class="stub"1
23
+…class="stub"1
2n
)  -class="stub"4-4n
2n+1
(9分)
=-4•
class="stub"1
22
(1-class="stub"1
2n-1
)
1-class="stub"1
2
-class="stub"4-4n
2n+1
=-2(1-class="stub"1
2n-1
)-class="stub"4-4n
2n+1
=class="stub"n+1
2n-1
-2
(11分)
Tn=class="stub"n+1
2n-2
-4
(12分)

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