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

题目简介

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

题目详情

数列{an}的前n项和为Sn,且a1=1,Sn+1=2Sn+n+1,n∈N.
(Ⅰ)求数列{an}的通项公式;
(Ⅱ)bn=
n
an+1-an
,设数列{bn}的前n项和为Tn,n∈N*,试判断Tn与2的关系,并说明理由.
题型:解答题难度:中档来源:不详

答案

(Ⅰ)∵a1=1,Sn+1=2Sn+n+1,
∴Sn+1+(n+1)+2=2(Sn+n+2),
并且S1+1+2=1+1+2=4,数列{Sn+n+1}组成一个以4为首项,2为公比的等比数列,
∴Sn+n+1=4×2n-1=2n+1,
Sn=2n+1-n-2.
∴a1=S1=22-1-2=1,
an=Sn-Sn-1
=(2n+1-n-2)-(2n-n-1)=2n-1,
当n=1时,2n-1=1=a1,
∴an=2n-1.
(Ⅱ)∵an=2n-1,
bn =class="stub"n
an+1-an
=class="stub"n
2 n+1-2n
=class="stub"n
2n

Tn=1×class="stub"1
2
+2×class="stub"1
2 2
+…+n×class="stub"1
2 n
,①
class="stub"1
2
Tn=1×class="stub"1
2 2
+2×class="stub"1
 3
+…+n×class="stub"1
2 n+1
,②
①-②,得class="stub"1
2
Tn=class="stub"1
2
+ class="stub"1
2 2
+class="stub"1
2 3
+…+class="stub"1
2 n
-n×class="stub"1
2 n+1

=
class="stub"1
2
(1-class="stub"1
2 n
)
1-class="stub"1
2
-n×class="stub"1
2 n+1

=1-class="stub"1
2 n
-class="stub"n
2 n+1

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

∴Tn<2.

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