An=根号Sn+根号(Sn+1) 证明根号Sn是等差数列,求An通项公式
人气:284 ℃ 时间:2019-10-19 20:56:27
解答
应该是 a(n+1)=[s(n)]^(1/2)+[s(n+1)]^(1/2)...s(n+1)-s(n)=a(n+1)=[s(n)]^(1/2)+[s(n+1)]^(1/2),{[s(n+1)]^(1/2)-[s(n)]^(1/2)}{[s(n+1)]^(1/2)+[s(n)]^(1/2)}=[s(n)]^(1/2)+[s(n+1)]^(1/2),[s(n+1)]^(1/2)-[s(n)]^...
推荐
- 数列an ,a1=1,当n>=2时,an=(根号sn+根号sn-1)/2,证明根号sn是等差数列,求an
- 等差数列an的前n项和为sn,a1=1+根号2,s3=9+3根号2
- 各项均为正数的数列An的前N项和为Sn,已知数列{根号Sn}是首项为1,公差为1的等差数列,求An的通项公式,
- 一直数列{an}中,a1=根号2/2,an=[2(Sn)^2]/(2Sn+1)(n>=2) 1、证明{1/Sn}是等差数列
- 已知等差数列an的前n项和为Sn,且对于任意的正整数n满足2根号下Sn=(an)+1
- We spent the whole evening talking about the things about the things and persons ( )none of us
- 三道数学填空题(在线1小时)急
- they____(have)the same hobby!
猜你喜欢