已知等差数列{an},公差d>0,前n项和为Sn,S3=6,且满足a3-a1,2a2,a8成等比数列.(Ⅰ)求{an}的通项公式;(Ⅱ)设bn=1an•an+2,求数列{bn}的前n项和Tn的值.-数

题目简介

已知等差数列{an},公差d>0,前n项和为Sn,S3=6,且满足a3-a1,2a2,a8成等比数列.(Ⅰ)求{an}的通项公式;(Ⅱ)设bn=1an•an+2,求数列{bn}的前n项和Tn的值.-数

题目详情

已知等差数列{an},公差d>0,前n项和为Sn,S3=6,且满足a3-a1,2a2,a8成等比数列.
(Ⅰ)求{an}的通项公式;
(Ⅱ)设bn=
1
anan+2
,求数列{bn}的前n项和Tn的值.
题型:解答题难度:中档来源:不详

答案

(Ⅰ)由S3=6,a3-a1,2a2,a8成等比数列,得
3a1+3d=6
4(a1+d)2=2d(a1+7d)
,即
a1+d=2
2a12+3a1d-5d2=0

解得:
a1=class="stub"10
3
d=-class="stub"4
3
a1=1
d=1

∵d>0,
a1=1
d=1

∴an=a1+(n-1)d=1+1×(n-1)=n;
(Ⅱ)bn=class="stub"1
anan+2
=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(n+1)
-class="stub"1
2(n+2)

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