数列{an}的前n项和Sn=n2an+b,若a1=12,a2=56.(1)求数列{an}的前n项和Sn;(2)求数列{an}的通项公式;(3)设bn=ann2+n-1,求数列{bn}的前n项和Tn.-

题目简介

数列{an}的前n项和Sn=n2an+b,若a1=12,a2=56.(1)求数列{an}的前n项和Sn;(2)求数列{an}的通项公式;(3)设bn=ann2+n-1,求数列{bn}的前n项和Tn.-

题目详情

数列{an}的前n项和Sn=
n2
an+b
,若a1=
1
2
a2=
5
6

(1)求数列{an}的前n项和Sn
(2)求数列{an}的通项公式;
(3)设bn=
an
n2+n-1
,求数列{bn}的前n项和Tn
题型:解答题难度:中档来源:浙江模拟

答案

(1)由S1=a1=class="stub"1
2
,得class="stub"1
a+b
=class="stub"1
2
,由S2=a1+a2=class="stub"4
3
,得class="stub"4
2a+b
=class="stub"4
3

a+b=2
2a+b=3
,解得
a=1
b=1
,故Sn=
n2
n+1
;               …(4分)
(2)当n≥2时,an=Sn-Sn-1=
n2
n+1
-
( n-1 )2
n
=
n3-( n-1 )2(n+1)
n(n+1)
=
n2+n-1
n2+n
.…(7分)
由于a1=class="stub"1
2
也适合an=
n2+n-1
n2+n
.                           …(8分)
an=
n2+n-1
n2+n
;                                         …(9分)
(3)bn=
an
n2+n-1
=class="stub"1
n( n+1 )
=class="stub"1
n
-class="stub"1
n+1
.                     …(10分)
∴数列{bn}的前n项和Tn=b1+b2+…+bn-1+bn=1-class="stub"1
2
+class="stub"1
2
-class="stub"1
3
+…+class="stub"1
n-1
-class="stub"1
n
+class="stub"1
n
-class="stub"1
n+1
=1-class="stub"1
n+1
=class="stub"n
n+1
.                                 …(14分)

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