已知数列{an}中,a1=1,且满足an+1=3an+1,n∈N,求数列{an}的(1)通项公式an(2)前n项和Sn.-数学

题目简介

已知数列{an}中,a1=1,且满足an+1=3an+1,n∈N,求数列{an}的(1)通项公式an(2)前n项和Sn.-数学

题目详情

已知数列{an}中,a1=1,且满足an+1=3an+1,n∈N,求数列{an}的
(1)通项公式an   
(2)前n项和Sn
题型:解答题难度:中档来源:不详

答案

(1)由an+1=3an+1得,an+1+class="stub"1
2
=3(an+class="stub"1
2
),
又a1+class="stub"1
2
=1+class="stub"1
2
=class="stub"3
2
,所以数列{an+class="stub"1
2
}各项不为0,
所以数列{an+class="stub"1
2
}是以class="stub"3
2
为首项、3为公比的等比数列,
所以an+class="stub"1
2
=class="stub"3
2
3n-1
=class="stub"1
2
3n

所以an=class="stub"1
2
(3n-1)

(2)由(1)得
Sn=a1+a2+…+an
=class="stub"1
2
(3-1)+class="stub"1
2
(32-1)
+…+class="stub"1
2
(3n-1)
=class="stub"1
2
[(3+32+…+3n)-n]
=class="stub"1
2
3(1-3n)
1-3
-class="stub"1
2
n

=class="stub"1
4
3n+1-class="stub"1
2
n-class="stub"3
4

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