已知数列{an}前n项和为Sn,且Sn=n2,(1)求{an}的通项公式(2)设bn=1anan+1,求数列{bn}的前n项和Tn.-数学

题目简介

已知数列{an}前n项和为Sn,且Sn=n2,(1)求{an}的通项公式(2)设bn=1anan+1,求数列{bn}的前n项和Tn.-数学

题目详情

已知数列{an}前 n项和为Sn,且Sn=n2
(1)求{an}的通项公式    
(2)设 bn=
1
anan+1
,求数列{bn}的前 n项 和Tn
题型:解答题难度:中档来源:不详

答案

(1)∵Sn=n2
∴Sn-1=(n-1)2
两个式子相减得
an=2n-1;                             
(2)bk=class="stub"1
akak+1
=class="stub"1
(2k-1)(2k+1)
=class="stub"1
2
(class="stub"1
2k-1
-class="stub"1
2k+1
)

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

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