已知等差数列{an}的公差d≠0,它的前n项和为Sn,若S5=70,且a2,a7,a22成等比数列.(1)求数列{an}的通项公式;(2)设数列{1Sn}的前n项和为Tn,求证:16≤Tn<38.-数

题目简介

已知等差数列{an}的公差d≠0,它的前n项和为Sn,若S5=70,且a2,a7,a22成等比数列.(1)求数列{an}的通项公式;(2)设数列{1Sn}的前n项和为Tn,求证:16≤Tn<38.-数

题目详情

已知等差数列{an}的公差d≠0,它的前n项和为Sn,若S5=70,且a2,a7,a22成等比数列.
(1)求数列{an}的通项公式;
(2)设数列{
1
Sn
}
的前n项和为Tn,求证:
1
6
Tn
3
8
题型:解答题难度:中档来源:惠州模拟

答案

(1)∵数列{an}是等差数列,
∴an=a1+(n-1)d,Sn=na1+
n(n-1)
2
d
.…(1分)
依题意,有
S5=70
a72=a2a22
5a1+10d=70
(a1+6d)2=(a1+d)(a1+21d).
…(3分)
解得a1=6,d=4.…(5分)
∴数列{an}的通项公式为an=4n+2(n∈N*).…(6分)
(2)证明:由(1)可得Sn=2n2+4n.…(7分)
class="stub"1
Sn
=class="stub"1
2n2+4n
=class="stub"1
2n(n+2)
=class="stub"1
4
class="stub"1
n
-class="stub"1
n+2
).…(8分)
∴Tn=class="stub"1
S1
+class="stub"1
S2
+class="stub"1
S3
+…+class="stub"1
Sn-1
+class="stub"1
Sn

=class="stub"1
4
[(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
)]…(9分)
=class="stub"1
4
(1+class="stub"1
2
-class="stub"1
n+1
-class="stub"1
n+2

=class="stub"3
8
-class="stub"1
4
class="stub"1
n+1
+class="stub"1
n+2
).…(10分)
∵Tn-class="stub"3
8
=-class="stub"1
4
class="stub"1
n+1
+class="stub"1
n+2
)<0,
∴Tn<class="stub"3
8
.…(11分)
∵Tn+1-Tn=class="stub"1
4
class="stub"1
n+1
-class="stub"1
n+3
)>0,所以数列{Tn}是递增数列.…(12分)
∴Tn≥T1=class="stub"1
6
.…(13分)
class="stub"1
6
≤Tn<class="stub"3
8
.…(14分)

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