在等差数列{an}中,a1=3,前n项和Sn满足条件Sn+2Sn=n+4n,n=1,2,3,…(Ⅰ)求数列{an}的通项公式;(Ⅱ)记bn=1Sn,数列{bn}的前n项和为Tn,求Tn.-数学

题目简介

在等差数列{an}中,a1=3,前n项和Sn满足条件Sn+2Sn=n+4n,n=1,2,3,…(Ⅰ)求数列{an}的通项公式;(Ⅱ)记bn=1Sn,数列{bn}的前n项和为Tn,求Tn.-数学

题目详情

在等差数列{an}中,a1=3,前n项和Sn满足条件
Sn+2
Sn
=
n+4
n
,n=1,2,3,…
(Ⅰ)求数列{an}的通项公式;
(Ⅱ)记bn=
1
Sn
,数列{bn}的前n项和为Tn,求Tn
题型:解答题难度:中档来源:不详

答案

(1)设等差数列{an}的公差为d,
Sn+2
Sn
=class="stub"n+4
n
对一切正自然数n都成立可知,
当n=1时,得:
S3
S1
=
3a1+3d
a1
=5
,又a1=3,所以d=2,
所以an=3+2(n-1)=2n+1.
(2)由(Ⅰ)知等差数列{an}的前n项和Sn=
n(3+2n+1)
2
=n(n+2)
bn=class="stub"1
Sn
=class="stub"1
n(n+2)
=class="stub"1
2
(class="stub"1
n
-class="stub"1
n+2
)

∴Tn=b1+b2+…+bn
=class="stub"1
2
(1-class="stub"1
3
+class="stub"1
2
-class="stub"1
4
+class="stub"1
3
-class="stub"1
5
+…+class="stub"1
n
-class="stub"1
n+2
)

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

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

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