已知数列{an}的前n项和Sn=n2+2n+3,(n∈N*)(1)求通项an;(2)求和1a1a2+1a2a3+1a3a4+…+1anan+1.-数学

题目简介

已知数列{an}的前n项和Sn=n2+2n+3,(n∈N*)(1)求通项an;(2)求和1a1a2+1a2a3+1a3a4+…+1anan+1.-数学

题目详情

已知数列{an}的前n项和Sn=n2+2n+3,(n∈N*)
(1)求通项an
(2)求和
1
a1a2
+
1
a2a3
+
1
a3a4
+…+
1
anan+1
题型:解答题难度:中档来源:不详

答案

(1)∵a1=S1=6,
∴当n≥2时,an=Sn-Sn-1=2n+1,
当n=1时,a1=3≠6,∴an=
6(n=1)
2n+1(n≥2).
…(6分)
(2)当n=1时,原式=class="stub"1
30

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
)

∴原式=class="stub"1
30
+class="stub"1
2
•(class="stub"1
5
-class="stub"1
7
+…+class="stub"1
2n+1
-class="stub"1
2n+3
)
=class="stub"1
30
+class="stub"1
2
(class="stub"1
5
-class="stub"1
2n+3
)
=class="stub"2
15
-class="stub"1
2(2n+3)
…(13分)

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