数列{an}中,a1=3,nan+1-(n+1)an=2n(n+1)(1)求证{ann}为等差数列,并求通项公式an;(2)设bn=(an-2n2)•3n,求数列{bn}的前n项和Sn.-数学

题目简介

数列{an}中,a1=3,nan+1-(n+1)an=2n(n+1)(1)求证{ann}为等差数列,并求通项公式an;(2)设bn=(an-2n2)•3n,求数列{bn}的前n项和Sn.-数学

题目详情

数列{an}中,a1=3,nan+1-(n+1)an=2n(n+1)
(1)求证{
an
n
}
为等差数列,并求通项公式an
(2)设bn=(an-2n2)•3n,求数列{bn}的前n项和Sn
题型:解答题难度:中档来源:不详

答案

证明:(1)由题意可得,
an+1
n+1
-
an
n
=2∴{
an
n
}
为等差数列.
an
n
=
a1
1
+2(n-1)=2n+1∴an=2n2+n

(2)由(1)可得,bn=n•3n
Sn=1•3+2•32+…n•3n
3Sn=1•32+2•33+…n•3n+1
∴-2Sn=3+32+…+3n-n•3n+1=
3(1-3n)
1-3
-n•3n+1

Sn=
3+(2n-1)•3n+1
4

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