在等差数列{an}中,a2+a3=7,a4+a5+a6=18.(1)求数列{an}的通项公式;(2)设数列{an}的前n项和为Sn,求1S3+1S6+…+1S3n.-数学

题目简介

在等差数列{an}中,a2+a3=7,a4+a5+a6=18.(1)求数列{an}的通项公式;(2)设数列{an}的前n项和为Sn,求1S3+1S6+…+1S3n.-数学

题目详情

在等差数列{an}中,a2+a3=7,a4+a5+a6=18.
(1)求数列{an}的通项公式;
(2)设数列{an}的前n项和为Sn,求
1
S3
+
1
S6
+…+
1
S3n
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,依题意,
a1+d+a1+2d=7
a1+3d+a1+4d+a1+5d=18
,解得a1=2,d=1,
∴an=2+(n-1)×1=n+1…5′
(2)S3n=
3n(a1+a3n)
2
=
3n(2+3n+1)
2
=
9n(n+1)
2

class="stub"1
S3n
=class="stub"2
9n(n+1)
=class="stub"2
9
class="stub"1
n
-class="stub"1
n+1
)…9′
class="stub"1
S3
+class="stub"1
S6
+…+class="stub"1
S3n
=class="stub"2
9
[(1-class="stub"1
2
)+(class="stub"1
2
-class="stub"1
3
)+…+(class="stub"1
n
-class="stub"1
n+1
)]=class="stub"2n
9(n+1)
…12′

更多内容推荐