已知数列{an}的前n项和,Sn=n2+2n+1.(1)求数列{an}的通项公式an;(2)记Tn=1a1a2+1a2a3+…+1anan+1,求Tn.-数学

题目简介

已知数列{an}的前n项和,Sn=n2+2n+1.(1)求数列{an}的通项公式an;(2)记Tn=1a1a2+1a2a3+…+1anan+1,求Tn.-数学

题目详情

已知数列{an}的前n项和,Sn=n2+2n+1
(1)求数列{an}的通项公式an
(2)记Tn=
1
a1a2
+
1
a2a3
+…+
1
anan+1
,求Tn
题型:解答题难度:中档来源:不详

答案

(I)当n=1时,a1=S1=4,
当n≥2时,an=Sn-Sn-1=n2+2n+1-[(n-1)2+2(n-1)+1]=2n+1,
又a1=4不适合上式,
an=
4,   n=1
2n+1,  n≥2

(II)∵class="stub"1
a1a2
=class="stub"1
4×5

当n≥2时,class="stub"1
anan+1
=class="stub"1
(2n+1)(2n+3)
=class="stub"1
2
(class="stub"1
2n+1
-class="stub"1
2n+3
)

Tn=class="stub"1
4×5
+class="stub"1
2
(class="stub"1
5
-class="stub"1
7
+class="stub"1
7
-class="stub"1
9
+…+class="stub"1
2n+1
-class="stub"1
2n+3
)

=class="stub"1
20
+class="stub"1
2
(class="stub"1
5
-class="stub"1
2n+3
)
=class="stub"3
20
-class="stub"1
2(2n+3)

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