问题 解答题
已知数列{an},{bn}中,a1=b1=1,且当n≥2时,an-nan-1=0,bn=2bn-1-2n-1.记n的阶乘n(n-1)(n-2)…3•2•1≈n!
(1)求数列{an}的通项公式;
(2)求证:数列{
bn
2n
}
为等差数列;
(3)若cn=
an
an+2
+bn-2n
,求{cn}的前n项和.
答案

(1)∵an-nan-1=0(n≥2),a1=1,

∴an=nan-1=n(n-1)an-2=n(n-1)(n-2)an-3=…

=n(n-1)(n-2)…3•2•1=n!

又a1=1=1!,∴an=n!

(2)证明:由bn=2bn-1-2n-1,两边同时除以2n得:

bn
2n
=
bn-1
2n-1
-
1
2
,即
bn
2n
-
bn-1
2n-1
=
1
2

∴数列{

bn
2n
}是以
1
2
为首项,公差为-
1
2
的等差数列,

bn
2n
=
1
2
+(n-1)(-
1
2
)=1-
n
2
,故bn=2n(1-
n
2
)

(3)因为

an
an+2
=
n!
(n+2)!
=
1
(n+1)(n+2)
=
1
n+1
-
1
n+2

bn-2n=2n(1-

n
2
)-2n=-n•2n-1

记An=

a1
a3
+
a2
a4
+
a3
a5
+…+
an
an+2

=(

1
2
-
1
3
)+(
1
3
-
1
4
)+(
1
4
-
1
5
)+…+(
1
n+1
-
1
n+2
)

=

1
2
-
1
n+2

记{bn-2n}的前n项和为Bn

Bn=-1•20-2•21-3•22-…-n•2n-1 ①

2Bn=-1•21-2•22-…-(n-1)•2n-1-n•2n ②

由②-①得:

Bn=20+21+22+…+2n-1-n•2n=

1-2n
1-2
-n•2n=(1-n)•2n-1.

∴Sn=c1+c2+c3+…+cn=An+Bn=(1-n)•2n-

1
2
-
1
n+2

所以数列{cn}的前n项和为(1-n)•2n-

1
2
-
1
n+2

填空题
问答题