Question

在数列{an}中,a1=2,an+1=4an-3n+1,n∈N*, (Ⅰ)证明:...

在数列{an}中,a1=2,an+1=4an-3n+1,n∈N*, (Ⅰ)证明:数列{an-n}是等比数列; (Ⅱ)设bn=nan-n2-n,求数列{bn}的前n项和Sn;

Answer 1

(Ⅰ)证明:由题设an+1=4an-3n+1,
得an+1-(n+1)=4(an-n),n∈N+
又a1-1=1≠0∴
an+1-(n+1)
an-n
=4
∴数列{an-n}是首项为1,且公比为4的等比数列
(Ⅱ)由(1)可知an-n=4n-1
而bn=n(an-n)-n=n•4n-1-n
∴Sn=1•40+2•41+3•42+n•4n-1-(1+2+3+n)Tn
=1•40+2•41+3•42+n•4n-1①
4Tn=1•41+2•42+3•43+(n-1)•4n-1+n•4n②
由①-②得:-3Tn=1+4+42+4n-1-n•4n=
1-4n
1-4
-n•4n=
4n-1
3
-n•4n
∴Tn=
1-4n
9
+
n•4n
3
=
1
9
+(
n
3
-
1
9
)•4n
=
(3n-1)•4n
9
+
1
9
=
(3n-1)•4n+1
9
Sn=
(3n-1)•4n+1
9
-
n(n+1)
2