设数列an的各项均为正数,它的前n项的和为Sn,满足Sn=1/8an²+1/2an+1/2;数列bn满足b1=a1,b(n+1)乘(a(n+1)-an)=bn,其中n∈N*.
(1)求数列an和bn的通项公式
(2)设Cn=an乘bn,求证:数列Cn的前n项和Tn>5/9 (n∈N*)
设数列an的各项均为正数,它的前n项的和为Sn,满足Sn=1/8an²+1/2an+1/2;数列b
答案:1 悬赏:0
解决时间 2021-02-27 22:56
- 提问者网友:野性
- 2021-02-27 13:04
最佳答案
- 二级知识专家网友:星星坠落
- 2021-02-27 13:13
Sn=1/8an²+1/2an+1/2 , bn=b(n+1)[(a(n+1)-an]
a1=a1=1/8a1²+1/2a1+1/2
a1=2
S(n-1)=1/8a(n-1)²+1/2a(n-1)+1/2
an=1/8an²-1/8a(n-1)²+1/2an-1/2a(n-1)
an²-a(n-1)²-4an-4a(n-1)=0
[an+a(n-1)][an-a(n-1)-4]=0
[an+a(n-1)]不为0
an-a(n-1)-4=0
an-a(n-1)=4
{an}是等差数列
an=2+4(n-1)=4n-2
an=4n-2
bn=b(n+1)[(a(n+1)-an]=4b(n+1)
b(n+1)/bn=1/4
{bn}是等比数列
bn=b1*(1/4)^(n-1)=8*(1/4)^n=2^(3-2n)
bn=2^(3-2n)
所以,an=4n-2, bn=2^(3-2n)
2)Cn=an*bn=(2n-1)*2^(4-2n)
Tn=4+3+5*(1/2)^2+7*(1/2)^4+...+(2n-3)*(1/2)^2(n-3)+(2n-1)*2^(4-2n)
(1/4)Tn=1+3*(1/2)^2+5*(1/2)^4+...+(2n-3)(1/2)^2(4-2n)+(2n-1)*2^(2-2n)
两式相减得
3/4Tn=4+2[1+(1/2)^2+(1/2)^4+...+(1/2)^(n-1)]-(2n-1)*2^(2-2n)
3/4Tn=4+2[4/3(1-(1/2)^(n-1)]-(2n-1)*2^(2-2n)
3/4Tn=4+8/3(1-(1/2)^(n-1)]-(2n-1)*2^(2-2n)
a1=a1=1/8a1²+1/2a1+1/2
a1=2
S(n-1)=1/8a(n-1)²+1/2a(n-1)+1/2
an=1/8an²-1/8a(n-1)²+1/2an-1/2a(n-1)
an²-a(n-1)²-4an-4a(n-1)=0
[an+a(n-1)][an-a(n-1)-4]=0
[an+a(n-1)]不为0
an-a(n-1)-4=0
an-a(n-1)=4
{an}是等差数列
an=2+4(n-1)=4n-2
an=4n-2
bn=b(n+1)[(a(n+1)-an]=4b(n+1)
b(n+1)/bn=1/4
{bn}是等比数列
bn=b1*(1/4)^(n-1)=8*(1/4)^n=2^(3-2n)
bn=2^(3-2n)
所以,an=4n-2, bn=2^(3-2n)
2)Cn=an*bn=(2n-1)*2^(4-2n)
Tn=4+3+5*(1/2)^2+7*(1/2)^4+...+(2n-3)*(1/2)^2(n-3)+(2n-1)*2^(4-2n)
(1/4)Tn=1+3*(1/2)^2+5*(1/2)^4+...+(2n-3)(1/2)^2(4-2n)+(2n-1)*2^(2-2n)
两式相减得
3/4Tn=4+2[1+(1/2)^2+(1/2)^4+...+(1/2)^(n-1)]-(2n-1)*2^(2-2n)
3/4Tn=4+2[4/3(1-(1/2)^(n-1)]-(2n-1)*2^(2-2n)
3/4Tn=4+8/3(1-(1/2)^(n-1)]-(2n-1)*2^(2-2n)
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