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{
(3)若cn=
|
(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
,即1 2
-bn 2n
=bn-1 2n-1
.1 2
∴数列{
}是以bn 2n
为首项,公差为-1 2
的等差数列,1 2
则
=bn 2n
+(n-1)(-1 2
)=1-1 2
,故bn=2n(1-n 2
).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-
)-2n=-n•2n-1.n 2
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.
则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=
-n•2n=(1-n)•2n-1.1-2n 1-2
∴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