正数数列{an}的前n项和为Sn,且2根号Sn=an+1 (1)求an (2)bn=1/(an×an+1),bn的前n项和为Tn,求Tn<1/2
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/02 02:16:24
![正数数列{an}的前n项和为Sn,且2根号Sn=an+1 (1)求an (2)bn=1/(an×an+1),bn的前n项和为Tn,求Tn<1/2](/uploads/image/z/634607-71-7.jpg?t=%E6%AD%A3%E6%95%B0%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E4%B8%942%E6%A0%B9%E5%8F%B7Sn%EF%BC%9Dan%EF%BC%8B1+%EF%BC%881%EF%BC%89%E6%B1%82an+%EF%BC%882%EF%BC%89bn%EF%BC%9D1%EF%BC%8F%EF%BC%88an%C3%97an%EF%BC%8B1%EF%BC%89%2Cbn%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BATn%2C%E6%B1%82Tn%EF%BC%9C1%EF%BC%8F2)
正数数列{an}的前n项和为Sn,且2根号Sn=an+1 (1)求an (2)bn=1/(an×an+1),bn的前n项和为Tn,求Tn<1/2
正数数列{an}的前n项和为Sn,且2根号Sn=an+1 (1)求an (2)bn=1/(an×an+1),
bn的前n项和为Tn,求Tn<1/2
正数数列{an}的前n项和为Sn,且2根号Sn=an+1 (1)求an (2)bn=1/(an×an+1),bn的前n项和为Tn,求Tn<1/2
(1)4Sn=(an +1)^2 4Sn-1=(an-1 +1)^2,两式相减得到4an=(an +1)^2-(an-1 +1)^2
移项整理,(an -1)^2=(an-1 +1)^2 开平方得到an=an-1 +2 或者an+an-1 =0(排除)
又n=1时a1=1,于是an=2n-1
(2)说实话你这+1到底是加在哪里根本不知道,是n上面还是an上面呢
bn=1/((2n-1)(2n+1))=1/2 * 2/((2n-1)(2n+1))=1/2*(1/(2n-1)-1/(2n+1))
Tn=b1+b2+...+bn=1/2*(1/1-1/3+1/3-1/5+...+1/(2n-1)-1/(2n+1))=1/2*(1-1/(2n+1))=n/(2n+1)