设数列{an}的前n项和为Sn,a1=1,且对任意正整数n,点(an+1,Sn)在直线2x+y-2=0上.(1)求数列{an}的通项公式;(2)若bn=nan2,求数列{bn}的前n项和.-数学

题目简介

设数列{an}的前n项和为Sn,a1=1,且对任意正整数n,点(an+1,Sn)在直线2x+y-2=0上.(1)求数列{an}的通项公式;(2)若bn=nan2,求数列{bn}的前n项和.-数学

题目详情

设数列{an}的前n项和为Sn,a1=1,且对任意正整数n,点(an+1,Sn)在直线2x+y-2=0上.
(1)求数列{an}的通项公式;
(2)若bn=nan2,求数列{bn}的前n项和.
题型:解答题难度:中档来源:不详

答案

(1):(Ⅰ)∵点(an+1,Sn)在直线2x+y-2=0上,
∴2an+1 +Sn -2=0. ①
当n≥2时,2an+sn-1-2=0. ②
①─②得 2an+1 -2an+an=0,即
an+1
an
=class="stub"1
2
(n≥2),
把n=1和a1=1代入①,可得a2=class="stub"1
2
,也满足上式,
∴{an}是首项为1,公比为class="stub"1
2
的等比数列,
则an=(class="stub"1
2
)n-1

(2)设数列{bn}的前n项和是Tn,
由(1)得,bn=nan2=n(class="stub"1
2
)
2(n-1)
=n(class="stub"1
4
)
n-1

∴Tn=1+2×class="stub"1
4
+3×class="stub"1
42
+…+n(class="stub"1
4
)
n-1
     ①,
class="stub"1
4
Tn
=class="stub"1
4
+2×class="stub"1
42
+3×class="stub"1
43
+…+n(class="stub"1
4
)
n
    ②,
①-②得,class="stub"3
4
Tn
=1+class="stub"1
4
+class="stub"1
42
+class="stub"1
43
+…+class="stub"1
4n-1
-n(class="stub"1
4
)
n

=
1-class="stub"1
4n
1-class="stub"1
4
-n(class="stub"1
4
)
n
=class="stub"3
4
(1-class="stub"1
4n
)-n(class="stub"1
4
)
n

则Tn=1-class="stub"4n+3
3•4n

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