已知数列{an}的前n项和为Sn,Sn+an=2-(12)n(n为正整数).(1)求数列{an}的通项公式;(2)若cnn+1=ann+2,Tn=c1+c2+…+cn,求Tn.-数学

题目简介

已知数列{an}的前n项和为Sn,Sn+an=2-(12)n(n为正整数).(1)求数列{an}的通项公式;(2)若cnn+1=ann+2,Tn=c1+c2+…+cn,求Tn.-数学

题目详情

已知数列{an}的前n项和为SnSn+an=2-(
1
2
)n
(n为正整数).
(1)求数列{an}的通项公式;
(2)若
cn
n+1
=
an
n+2
,Tn=c1+c2+…+cn,求Tn
题型:解答题难度:中档来源:不详

答案

(1)∵Sn+an=2-(class="stub"1
2
)
n
,∴n≥2时,Sn-1+an-1=2-(class="stub"1
2
)
n-1

两式相减可得2an-an-1=(class="stub"1
2
)n

2n+1an-2nan-1=1
∵n=1时,S1+a1=2-(class="stub"1
2
)
1
,∴a1=class="stub"3
4
,∴22a1=3
∴{2n+1an}是以3为首项,1为公差的等差数列,
∴2n+1an=n+2,∴an=class="stub"n+2
2n+1

(2)∵
cn
n+1
=
an
n+2
,∴cn=class="stub"n+1
n+2
an
=class="stub"n+1
n+2
×class="stub"n+2
2n+1
=class="stub"n+1
2n+1

∴Tn=c1+c2+…+cn=class="stub"2
22
+class="stub"3
23
+class="stub"4
24
+…+class="stub"n
2n
+class="stub"n+1
2n+1

2Tn=class="stub"2
2
+class="stub"3
22
+class="stub"4
23
+…+class="stub"n+1
2n

两式相减得Tn=class="stub"2
2
+class="stub"1
22
+class="stub"1
23
+…class="stub"1
2n
-class="stub"n+1
2n+1
=class="stub"1
2
+class="stub"1
2
+class="stub"1
22
+…+class="stub"1
2n
-class="stub"n+1
2n+1

=class="stub"1
2
+
class="stub"1
2
(1-class="stub"1
2n
)
1-class="stub"1
2
-class="stub"n+1
2n+1

=class="stub"3
2
-class="stub"2+n
2n+1

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