设数列{an}的前n项和为Sn,已知a1=1,Sn=2n+1-n-2(n∈N*),(Ⅰ)求数列{an}的通项公式;(Ⅱ)若bn=nan+1-an,数列{bn}的前项和为Tn.-数学

题目简介

设数列{an}的前n项和为Sn,已知a1=1,Sn=2n+1-n-2(n∈N*),(Ⅰ)求数列{an}的通项公式;(Ⅱ)若bn=nan+1-an,数列{bn}的前项和为Tn.-数学

题目详情

设数列{an}的前n项和为Sn,已知a1=1,Sn=2n+1-n-2(n∈N*),
(Ⅰ)求数列{an}的通项公式;
(Ⅱ)若bn=
n
an+1-an
,数列{bn}的前项和为Tn
题型:解答题难度:中档来源:不详

答案

(Ⅰ)∵Sn=2n+1-n-2
当n≥2时Sn-1=2n-(n-1)-2(n∈N*)
∴an=2n-1(n≥2)
又a1=S1=1
∴an=2n-1(n∈N*)
(Ⅱ)∵an=2n-1∴bn=class="stub"n
(2n+1-1)-(2n-1)
=class="stub"n
2n+1-2n
=class="stub"n
2n

Tn=class="stub"1
2
+class="stub"2
22
+class="stub"3
23
+…+class="stub"n
2n
class="stub"1
2
Tn=,class="stub"1
22
+class="stub"2
23
+…+class="stub"n-1
2n
+class="stub"n
2n+1

Tn=2(class="stub"1
2
+class="stub"1
22
+class="stub"1
23
+…+class="stub"1
2n
-class="stub"n
2n+1
)=2-class="stub"1
2n-1
-class="stub"n
2n

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