数学试题:已知数列{an}前n项和为Sn
来源:学生作业帮 编辑:百度作业网作业帮 分类:数学作业 时间:2024/07/08 16:11:35
数学试题:已知数列{an}前n项和为Sn
已知数列{an}前n项和为Sn,且满足Sn=1-n×an(n=1,2,3...).
(I)求a1、a2的值;
(II)求an.
谢谢您.
已知数列{an}前n项和为Sn,且满足Sn=1-n×an(n=1,2,3...).
(I)求a1、a2的值;
(II)求an.
谢谢您.
![数学试题:已知数列{an}前n项和为Sn](/uploads/image/z/5689321-25-1.jpg?t=%E6%95%B0%E5%AD%A6%E8%AF%95%E9%A2%98%EF%BC%9A%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn)
S1=a1=1-1*a1
2a1=1 a1=1/2
S2=1-2a2=a1+a2=1/2+a2
3a2=1/2 a2=1/6
Sn=1-nan
Sn-1=1-(n-1)a(n-1)
相减
an=Sn-Sn-1=1-nan-1+(n-1)a(n-1)
an=(n-1)a(n-1)/(n+1)
=(n-1)(n-2)a(n-2)/[n(n+1)]
=(n-1)(n-2)(n-3)a(n-3)/[n(n+1)(n-1)]
=...
=[(n-1)(n-2)...1]a1/[(n+1)n(n-1)...3]
=1*2a1/[n(n+1)]
=1/[n(n+1)]
an=1/[n(n+1)]=1/n-1/(n+1)
2a1=1 a1=1/2
S2=1-2a2=a1+a2=1/2+a2
3a2=1/2 a2=1/6
Sn=1-nan
Sn-1=1-(n-1)a(n-1)
相减
an=Sn-Sn-1=1-nan-1+(n-1)a(n-1)
an=(n-1)a(n-1)/(n+1)
=(n-1)(n-2)a(n-2)/[n(n+1)]
=(n-1)(n-2)(n-3)a(n-3)/[n(n+1)(n-1)]
=...
=[(n-1)(n-2)...1]a1/[(n+1)n(n-1)...3]
=1*2a1/[n(n+1)]
=1/[n(n+1)]
an=1/[n(n+1)]=1/n-1/(n+1)
数学试题:已知数列{an}前n项和为Sn
已知数列{an}的前n项和为Sn
已知数列{an}前n项和为Sn,且Sn=-2an+3
一道关于数列 已知数列{An}的前n项和为Sn,Sn=3+2An,求An
已知数列{an}的前n项和为Sn,且Sn=n-5an-85,n∈N*
已知数列{an}的前n项和为Sn,且Sn=n-5an-85,n∈N*
已知数列{an}的前n项和为Sn,且Sn=n-5an-85,n属于正整数
已知数列{an}的前n项和为Sn,Sn=(an-1)/3 (n∈N)
已知数列{an}的前n项和为Sn,Sn=13(an−1)(n∈N*).
已知数列{an}的前n项和为Sn,且Sn=23an+1(n∈N*);
已知数列{An}的前n项和为Sn,且Sn=n²+n(n∈N*)
已知数列 an前n项和为Sn,a1=1,Sn=2a(n+1),求Sn