设等差数列{an}的前n项和为Sn,已知a3=9,S6=66.(1)求数列{an}的通项公式an及前n项的和Sn;(2)设数列{1anan+1}的前n项和为Tn,证明:Tn<14.-高二数学

题目简介

设等差数列{an}的前n项和为Sn,已知a3=9,S6=66.(1)求数列{an}的通项公式an及前n项的和Sn;(2)设数列{1anan+1}的前n项和为Tn,证明:Tn<14.-高二数学

题目详情

设等差数列{an}的前n项和为Sn,已知a3=9,S6=66.
(1)求数列{an}的通项公式an及前n项的和Sn
(2)设数列{
1
anan+1
}
的前n项和为Tn,证明:Tn
1
4
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,
由题意可得
a3=a1+2d=9
S6=6a1+class="stub"6×5
2
d=66

解之可得a1=1,d=4,故an=1+4(n-1)=4n-3,
所以Sn=
n(a1+an)
2
=
n(1+4n-3)
2
=2n2-n;
(2)由(1)可知class="stub"1
anan+1
=class="stub"1
(4n-3)(4n-1)
=class="stub"1
4
class="stub"1
4n-3
-class="stub"1
4n+1
),
故Tn=class="stub"1
4
[(1-class="stub"1
5
)+(class="stub"1
5
-class="stub"1
9
)+…+(class="stub"1
4n-3
-class="stub"1
4n+1
)]
=class="stub"1
4
(1-class="stub"1
4n+1
)=class="stub"n
4n+1
class="stub"n
4n
=class="stub"1
4
,命题得证.

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