已知等差数列{an}的首项a1=3,公差d≠0,其前n项和为Sn,且a1,a4,a13成等比数列.(Ⅰ)求数列{an}的通项公式;(Ⅱ)证明:1S1+1S2+…1Sn<34.-数学

题目简介

已知等差数列{an}的首项a1=3,公差d≠0,其前n项和为Sn,且a1,a4,a13成等比数列.(Ⅰ)求数列{an}的通项公式;(Ⅱ)证明:1S1+1S2+…1Sn<34.-数学

题目详情

已知等差数列{an}的首项a1=3,公差d≠0,其前n项和为Sn,且a1,a4,a13成等比数列.
(Ⅰ)求数列{an}的通项公式;
(Ⅱ)证明:
1
S1
+
1
S2
+…
1
Sn
3
4
题型:解答题难度:中档来源:不详

答案

由a1,a4,a13成等比数列,得a42=a1a13
(a1+3d)2=a1(a1+12d),所以a12+6a1d+9d2=a12+12a1d
9d2=6a1d,a1=class="stub"3
2
d
.则d=class="stub"2
3
a1=class="stub"2
3
×3=2

(Ⅰ)an=a1+(n-1)d=3+2(n-1)=2n+1;
(Ⅱ)Sn=na1+
n(n-1)d
2
=3n+n2-n=n(n+2)

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

所以class="stub"1
S1
+class="stub"1
  S2
+…class="stub"1
Sn
=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-1
-class="stub"1
n+1
+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)
class="stub"3
4

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