数列an的前n项和为Sn,且满足an+2Sn*S(n-1)=0,(n大于等于2),a1=1/2.证1/Sn等差,求an表达式!希望速...数列an的前n项和为Sn,且满足an+2Sn*S(n-1)=0,(n大于等于2),a1=1/2.证1/Sn等差,求an表达式!
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/30 06:50:33
![数列an的前n项和为Sn,且满足an+2Sn*S(n-1)=0,(n大于等于2),a1=1/2.证1/Sn等差,求an表达式!希望速...数列an的前n项和为Sn,且满足an+2Sn*S(n-1)=0,(n大于等于2),a1=1/2.证1/Sn等差,求an表达式!](/uploads/image/z/1094015-47-5.jpg?t=%E6%95%B0%E5%88%97an%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E4%B8%94%E6%BB%A1%E8%B6%B3an%2B2Sn%2AS%28n-1%29%3D0%2C%28n%E5%A4%A7%E4%BA%8E%E7%AD%89%E4%BA%8E2%29%2Ca1%3D1%2F2.%E8%AF%811%2FSn%E7%AD%89%E5%B7%AE%2C%E6%B1%82an%E8%A1%A8%E8%BE%BE%E5%BC%8F%21%E5%B8%8C%E6%9C%9B%E9%80%9F...%E6%95%B0%E5%88%97an%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E4%B8%94%E6%BB%A1%E8%B6%B3an%2B2Sn%2AS%28n-1%29%3D0%2C%28n%E5%A4%A7%E4%BA%8E%E7%AD%89%E4%BA%8E2%29%2Ca1%3D1%2F2.%E8%AF%811%2FSn%E7%AD%89%E5%B7%AE%2C%E6%B1%82an%E8%A1%A8%E8%BE%BE%E5%BC%8F%21)
数列an的前n项和为Sn,且满足an+2Sn*S(n-1)=0,(n大于等于2),a1=1/2.证1/Sn等差,求an表达式!希望速...数列an的前n项和为Sn,且满足an+2Sn*S(n-1)=0,(n大于等于2),a1=1/2.证1/Sn等差,求an表达式!
数列an的前n项和为Sn,且满足an+2Sn*S(n-1)=0,(n大于等于2),a1=1/2.证1/Sn等差,求an表达式!希望速...
数列an的前n项和为Sn,且满足an+2Sn*S(n-1)=0,(n大于等于2),a1=1/2.证1/Sn等差,求an表达式!
数列an的前n项和为Sn,且满足an+2Sn*S(n-1)=0,(n大于等于2),a1=1/2.证1/Sn等差,求an表达式!希望速...数列an的前n项和为Sn,且满足an+2Sn*S(n-1)=0,(n大于等于2),a1=1/2.证1/Sn等差,求an表达式!
an+2Sn*S(n-1)=0
而an=Sn-S(n-1)
∴Sn-S(n-1)+2Sn*S(n-1)=0
同除以Sn*S(n-1)整理:
1/Sn -1/S(n-1)=2
∴{1/Sn}为等差数列,公差2,首项=1/a1=2
1/Sn=2+2(n-1)=2n
Sn=1/(2n)
an=Sn-S(n-1)=1/(2n)-1/(2n-2)
已知数列{an}的前n项和为Sn,且满足Sn=2an-1(n属于正整数),求数列{an}的通项公式an
已知数列{an}的前n项和为sn,且满足sn=n
已知数列{an}的前n项和为Sn,且满足Sn=2an-1,n为正整数,求数列{an}的通项公式anRT ,
已知数列{an}a1=2前n项和为Sn 且满足Sn Sn-1=3an 求数列{an}的通项公式an已知数列{an}a1=2前n项和为Sn 且满足Sn +Sn-1=3an 求数列{an}的通项公式an
已知正项数列an的前n项和为sn,且满足:an平方=2sn-an(n属于N*).求an的通项公式;2.求数列{an,2an(此an
已知数列an的前n项和为Sn,且满足3an=3+2Sn.求数列an通项公式?
三校生数学,an和sn的关系.已知数列{an}中,an>0,前n项之和为An,且满足An三校生数学,an和sn的关系.已知数列{an}中,an>0,前n项之和为An,且满足An=1/8(an+2)²,求数列{an}
已知数列 {an} 的前n项和为 Sn,且满足 Sn=3/2(an-1) (n∈正整数) 求 an 的通项公式
已知数列{an}的前n项和为Sn,且满足an+2Sn*Sn-1=0,a1=1/2.求证:{1/Sn}是等差数列
已知正数数列{an}的前n项和为Sn,且对于任意正整数n满足2根号Sn=an+1 求an通项
已知正数数列{an}的前n项和为Sn,且对于任意正整数n满足2根号Sn=an+1 求an通项
已知正数数列{an}的前n项和为Sn,且对于任意正整数n满足2根号Sn=an+1 求an通项
已知数列{an}中,an>0,前n项和为Sn,且满足Sn=1/8(an+2)^2.求证数列{an}是等差数列.
已知正整数数列{an}中,其前n项和为sn,且满足Sn=1/8(an+2)2求{an}的通项公式
已知数列{an}的前n项和为sn,且满足sn=1/2(an+1).求通项公式an.
已知数列{an}的首项a1=1/2,Sn是其前n项的和,且满足Sn=n^2an,则次数列的通项公式为an=?Sn=n²an
【高考】若数列{an}满足,a1=1,且a(n+1)=an/(1+an),设数列{bn}的前n项和为Sn,且Sn=2-bn,求{bn/an}的前...【高考】若数列{an}满足,a1=1,且a(n+1)=an/(1+an),设数列{bn}的前n项和为Sn,且Sn=2-bn,求{bn/an}的前n项和Tn
已知正项数列{an}的前n项和为sn,且满足sn+sn-1=kan^2+2 求an