数列{AN}满足a1=2,An+1=an2+6an+61.求数列{AN}的通项公式2.设bn=1/(an-6)-1/(an2+6an),{BN}前N项和为TN,求证:-5/16
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/11 18:26:10
数列{AN}满足a1=2,An+1=an2+6an+6
1.求数列{AN}的通项公式
2.设bn=1/(an-6)-1/(an2+6an),{BN}前N项和为TN,
求证:-5/16
(1)a(n+1)=(an)^2+6an+6 两边同加3得到
a(n+1)+3=(an)^2+6an+9=(an+3)^2;
所以an+3=[a(n-1)+3]^2=[a(n-2)+3]^4=……=(a1+3)^(2^(n-1))=5^(2^(n-1));
所以 an=5^(2^(n-1))-3;
(2)bn=1/(an-6)-1/(an2+6an)
=1/(an-6)-1/(a(n+1)-6);
所以
Tn=1/(a1-6)-1/(a2-6)+1/(a2-6)-1/(a3-6)+…+1/(an-6)-1/(a(n+1)-6)
=1/(a1-6)-1/(a(n+1)-6) 代入a1=2 和 a(n+1)=5^(2^n)-3;
= -1/4-1/[5^(2^n)-9];
显然对任意自然数1/[5^(2^n)-9]>0,所以Tn< -1/4;
又因为[5^(2^n)-9]单增,即1/[5^(2^n)-9]单减,故-1/[5^(2^n)-9]也单增,所以Tn在n=1时取最小值,即
Tn>=T1= -1/4-1/16= -5/16
数列{An}满足a1=1/2,a1+a2+..+an=n方an,求an
数列{an}满足a1=1 an+1=2n+1an/an+2n
数列an满足a1=2,an+1=4an+9,则an=?
数列{an}满足a1=1,且an=an-1+3n-2,求an
已知数列{an}满足an+1=2an+3.5^n,a1=6.求an
数列an满足a1=2,an+1=an²求an
数列an满足a1=2,an+1=an²求an
已知数列{an}满足a1=1 an+1=an/(3an+1) 则球an
已知数列an满足 a1=1/2,an+1=3an/an+3求证1/an为等差数列已知数列an满足 a1=1/2,an+1=3an/an+3求证1/an为等差数列
已知数列{an}满足a1=1,a2=3,an+2=3an+1-2an求an
已知数列an满足a1=1,1/an+1=根号1/an^2+2,an>0,求an
数列{an}满足:a1=1,an>0,an+1^2-an^2=1,那么an
数列{an}满足:a1=1,an>0,an+1^2-an^2=1,那么an
知数列{an}满足a1=-2,an+1(下标)=2+2an/(1-an),则an为?
已知数列{an}满足a1=1,an+1 -an+2an+1•an=0求通项
已知数列{an}满足a1=2,an+1=2an/an+2,则an等于多少
已知数列{an}满足a1=2,an+1-an=an+1*an,那么a31等于
数列an满足,a1=8且8an+1an-16an+1+2an+5=0