(文){an}是等差数列,公差d>0,Sn是{an}的前n项和.已知a1a4=22.S4=26.(1)求数列{an}的通项公式an;(2)令bn=1anan+1,求数列{bn}前n项和Tn.-数学

题目简介

(文){an}是等差数列,公差d>0,Sn是{an}的前n项和.已知a1a4=22.S4=26.(1)求数列{an}的通项公式an;(2)令bn=1anan+1,求数列{bn}前n项和Tn.-数学

题目详情

(文) {an}是等差数列,公差d>0,Sn是{an}的前n项和.已知a1a4=22.S4=26.
(1)求数列{an}的通项公式an
(2)令bn=
1
anan+1
,求数列{bn}前n项和Tn
题型:解答题难度:中档来源:宁波模拟

答案

(1)因为S4=
4(a1+a2)
2
=2(a1+a4)=26,得a1+a4=13  ①
又a1•a4=22  ②
由①得a4=13-a1 代入②得a1(13-a1)=22
解得a1=11或a1=2
a1=11时,a4=2,d<0不合题意,舍去
所以a1=2,a4=2+3d=11
d=3
所以an=2+3(n-1)=3n-1
(2)bn=class="stub"1
anan+1

Tn=class="stub"1
a1a2
+ class="stub"1
a2a3
 +class="stub"1
a3a4
+…+ class="stub"1
anan+1

因为class="stub"1
anan+1
= (class="stub"1
an
- class="stub"1
an+1
)(class="stub"1
an+1-an
)

因为an+1-an=d
所以class="stub"1
anan+1
= (class="stub"1
an
-class="stub"1
an+1
)•class="stub"1
3

Tn=class="stub"1
3
[class="stub"1
a1
-class="stub"1
a2
+class="stub"1
a2
-class="stub"1
a3
+ …+class="stub"1
an
-class="stub"1
an+1
]
=class="stub"1
3
×[class="stub"1
a1
-class="stub"1
an+1
]

=class="stub"1
3
×[class="stub"1
2
-class="stub"1
3n+2
]

=class="stub"n
6n+4

所以Tn=class="stub"n
6n+4

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