在数列{an}中,a1=1,an+1=(1+1n)an+n+12n.(1)设bn=ann,求数列{bn}的通项公式;(2)求数列{an}的前n项和Sn.-数学

题目简介

在数列{an}中,a1=1,an+1=(1+1n)an+n+12n.(1)设bn=ann,求数列{bn}的通项公式;(2)求数列{an}的前n项和Sn.-数学

题目详情

在数列{an}中,a1=1,an+1=(1+
1
n
)an+
n+1
2n

(1)设bn=
an
n
,求数列{bn}的通项公式;
(2)求数列{an}的前n项和Sn
题型:解答题难度:中档来源:不详

答案

(1)由已知得b1=a1=1,且
an+1
n+1
=
an
n
+class="stub"1
2n

即bn+1=bn+class="stub"1
2n
,从而b2=b1+class="stub"1
2

b3=b2+class="stub"1
22

bn=bn-1+class="stub"1
2n-1
(n≥2).
于是bn=b1+class="stub"1
2
+class="stub"1
22
+…+class="stub"1
2n-1
=2-class="stub"1
2n-1
(n≥2).
又b1=1,
故所求的通项公式为bn=2-class="stub"1
2n-1

(2)由(1)知an=2n-class="stub"n
2n-1

故Sn=(2+4++2n)-(1+class="stub"2
2
+class="stub"3
22
+class="stub"4
23
+…+class="stub"n
2n-1
),
设Tn=1+class="stub"2
21
+class="stub"3
22
+class="stub"4
23
+…+class="stub"n
2n-1
,①
class="stub"1
2
Tn=class="stub"1
2
+class="stub"2
22
+class="stub"3
23
+…+class="stub"n-1
2n-1
+class="stub"n
2n
,②
①-②得,
class="stub"1
2
Tn=1+class="stub"1
2
+class="stub"1
22
+class="stub"1
23
+…+class="stub"1
2n-1
-class="stub"n
2n

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

∴Tn=4-class="stub"n+2
2n-1

∴Sn=n(n+1)+class="stub"n+2
2n-1
-4.

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