数列{an}是等差数列,a2=3,前四项和S4=16.(1)求数列{an}的通项公式;(2)记Tn=1a1a2+1a2a3+…+1anan+1,计算T2011.-数学

题目简介

数列{an}是等差数列,a2=3,前四项和S4=16.(1)求数列{an}的通项公式;(2)记Tn=1a1a2+1a2a3+…+1anan+1,计算T2011.-数学

题目详情

数列{an}是等差数列,a2=3,前四项和S4=16.
(1)求数列{an}的通项公式;
(2)记Tn=
1
a 1a2
+
1
a2a3
+…+
1
anan+1
,计算T2011
题型:解答题难度:中档来源:不详

答案

(1)由a2=3,S4=16,根据题意得:
a1+d=3①
4a1+6d=16②
,解得:
a1=1
d=2

则an=1+2(n-1)=2n-1;
(2)∵class="stub"1
anan+1
=class="stub"1
(2n-1)(2n+1)
=class="stub"1
2
class="stub"1
2n-1
-class="stub"1
2n+1
),
∴T2011=class="stub"1
a1a2
+class="stub"1
a2a3
+…+class="stub"1
a2011a2012

=class="stub"1
1×3
+class="stub"1
3×5
+…+class="stub"1
2009×2011
+class="stub"1
2011×2013
+…+class="stub"1
4021×4023

=class="stub"1
2
(1-class="stub"1
3
+class="stub"1
3
-class="stub"1
5
+…+class="stub"1
2011
-class="stub"1
2013
+…+class="stub"1
4021
-class="stub"1
4023

=class="stub"1
2
(1-class="stub"1
4023

=class="stub"2011
4023

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