Question

高一数学数列题

已知数列an/n*2是公比为1/2的等比数列，a1=1 1、求an 2、令bn=a(n+1)-1/2(an),数列bn的前n项和为sn，证明sn<5
急！

Answer 1

an/n*2是公比为1/2的等比数列
a1/1^2=1
所以an/n^2=1*(1/2)^n-1)
1. an=n^2/2^(n-1)
2. bn=a(n+1)-(1/2)an
=(n+1)^2/2^n-1/2^n
=n/2^n
Sn=1/2+2/2^2+....+n/2^n
2Sn=1+2/2+3/2^2+....+n/2^(n-1)
后式减前式 Sn=1+1/2+1/2^2+....+1/2^(n-1)-n/2^n
=[1-(1/2)^n]/(1-1/2)-n/2^n
=2-2/2^n-n/2^n
=2-(n+1)/2^n
故Sn<2
也即Sn<5
证毕

Answer 2

1.数列{an/n^2}是公比为1/2的等比数列，a1=1 ,
∴an/n^2=(1/2)^(n-1),
∴an=n^2*(1/2)^(n-1).

2.bn=a<n+1>-(1/2)an
=(n+1)^2*(1/2)^n-n^2*(1/2)^n
=(1/2)^n*[(n+1)^2-n^2]
=(1/2)^n*(1+2n),
∴Sn=3/2+5/4+7/8+……+（1+2n)/2^n,
(1/2)Sn=..3/4+5/8+……+（-1/2n)/2^n+(1+2n)/2^(n+1),
相减得
（1/2)Sn=3/2+2/4+2/8+……+2/2^n-(1+2n)/2^(n+1)
=1/2+2-1/2^(n-1)-(1+2n)/2^(n+1),
∴Sn=5-1/2^(n-2)-(1+2n)/2^n<5.

Answer 3

①∵a1/1²=1
∴an/n²=(1/2)^(n-1)
∴an=n²/2^(n-1)
②bn=a(n+1)-1/2(an)=[(n+1)²-n²]/2^n=(2n+1)/2^n
Sn=3/2+5/2²+…+(2n+1)/2^n
1/2Sn=3/2²+5/2³+…+(2n+1)/2^(n+1)
用错位相减法得1/2Sn=3/2+2/2²+2/2³+…+2/2^n-(2n+1)/2^(n+1)=3/2+1/2+1/2²+…+1/2^(n-1)-(2n+1)/2^(n+1)=3/2+1-(1/2)^(n-1)-(2n+1)/2^(n+1)=5/2-(2n+5)/2^(n+1)
得Sn=5-(2n+5)/2^n<5

Answer 4

1、令cn=an/n*2,c1=a1/1*2=1,q=1/2
cn=1*(1/2)^(n-1)
an=cn*n^2=n^2/2^(n-1)