已知在数列{an}中,Sn是前n项和,满足Sn+an=n,(n=1,2,3,…).(Ⅰ)求a1,a2,a3的值;(Ⅱ)求数列{an}的通项公式;(Ⅲ)令bn=(2-n)(an-1)(n=1,2,3,…

题目简介

已知在数列{an}中,Sn是前n项和,满足Sn+an=n,(n=1,2,3,…).(Ⅰ)求a1,a2,a3的值;(Ⅱ)求数列{an}的通项公式;(Ⅲ)令bn=(2-n)(an-1)(n=1,2,3,…

题目详情

已知在数列{an}中,Sn是前n项和,满足Sn+an=n,(n=1,2,3,…).
(Ⅰ)求a1,a2,a3的值;
(Ⅱ)求数列{an}的通项公式;
(Ⅲ)令bn=(2-n)(an-1)(n=1,2,3,…),求数列{bn}的前n项和Tn
题型:解答题难度:中档来源:河北区一模

答案

(Ⅰ)∵Sn+an=n,
∴a1=class="stub"1
2
,a2=class="stub"3
4
,a3=class="stub"7
8
.…(3分)
(Ⅱ)∵a1+a2+a3+…+an-1+an=n-an,…①
a1+a2+a3+…+an+an+1=n+1-an+1,…②
②-①得2an+1-an=1,
即an+1-1=class="stub"1
2
(an-1).…(5分)
又a1-1=-class="stub"1
2

∴数列{an-1}是以-class="stub"1
2
为首项,以class="stub"1
2
为公比的等比数列.…(7分)
∴an-1=(-class="stub"1
2
(class="stub"1
2
)
n-1
=-(class="stub"1
2
)
n

可得an=1-(class="stub"1
2
)
n
.…(8分)
(Ⅲ)由(Ⅱ)知,an=1-(class="stub"1
2
)
n

∵bn=(2-n)(an-1)=(2-n)[-(class="stub"1
2
)
n
]=class="stub"n-2
2n
,…(10分)
所以数列{bn}的前n项和Tn=class="stub"-1
2
+class="stub"0
22
+class="stub"1
23
+…+class="stub"n-2
2n
.…①
所以class="stub"1
2
Tn=class="stub"-1
22
+class="stub"0
23
+class="stub"1
24
+…+class="stub"n-2
2n+1
.…②…(12分)
①-②得class="stub"1
2
Tn=class="stub"-1
2
+class="stub"1
22
+class="stub"1
23
+…+class="stub"1
2n
-class="stub"n-2
2n+1
=-class="stub"n
2n+1

所以Tn=-class="stub"n
2n
.…(14分)

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