求解答网！An=n²-1怎么来的

来源：蜘蛛抓取(WebSpider) 时间：2015-08-10 06:04 标签： an 1 2 n an 4求通项
为什么“a²(n+1)=an*a(n+2)”不能证明｛an}是等比数列_作业帮
为什么“a²(n+1)=an*a(n+2)”不能证明｛an}是等比数列
为什么“a²(n+1)=an*a(n+2)”不能证明｛an}是等比数列
因为你不能判断an或a(n+1) 是否非零 ,假设 a1=1 a2=0 a3=0,a4=0,a5=0.,这不是等比数列 但满足条件已知数列{an}的前n项和为Sn,a1=1,且2n（Sn+1）－2（n+1）Sn,=n²+n（n∈N*）求数列{an}的通项公式?_作业帮
已知数列{an}的前n项和为Sn,a1=1,且2n（Sn+1）－2（n+1）Sn,=n²+n（n∈N*）求数列{an}的通项公式?
已知数列{an}的前n项和为Sn,a1=1,且2n（Sn+1）－2（n+1）Sn,=n²+n（n∈N*）求数列{an}的通项公式?
2nS(n+1)-2(n+1)Sn=n^2+n=n(n+1)2S(n+1)/(n+1)-2Sn/n=12S1/1=2a1=2所以,2Sn/n=2、Sn=n.n>=2时,an=Sn-S(n-1)=n-(n-1)=1,n=1时也适合此式.所以,通项公式为an=1,其中n为正整数.已知数列{an}满足：a1+8/7*a2+（8/7）²*a3+...（8/7）^(n-1)an=n（n+1）,n∈N+,求an_作业帮
已知数列{an}满足：a1+8/7*a2+（8/7）²*a3+...（8/7）^(n-1)an=n（n+1）,n∈N+,求an
已知数列{an}满足：a1+8/7*a2+（8/7）²*a3+...（8/7）^(n-1)an=n（n+1）,n∈N+,求an
{a1+8/7*a2+（8/7）²*a3+...（8/7）^(n-1)an}-{a1+8/7*a2+（8/7）²*a3+...（8/7）^(n-2)a（n-1）}=n（n+1）-（n-1）n=2n即有（8/7）^(n-1)an=2nan=2n(7/8)^(n-1)The resource cannot be found.
Server Error in '/' Application.
The resource cannot be found.
Description: HTTP 404. The resource you are looking for (or one of its dependencies) could have been removed, had its name changed, or is temporarily unavailable. &Please review the following URL and make sure that it is spelled correctly.
Requested URL: /Detail_102483.aspx{an}的前n项和Sn=(3n²+5n)/(n²+3n+2),求lim(n→+∞)[a(n+1)/an]的值等待妙招有刚看到题的不要管网友推荐,我会取消网友推荐的_作业帮
{an}的前n项和Sn=(3n²+5n)/(n²+3n+2),求lim(n→+∞)[a(n+1)/an]的值等待妙招有刚看到题的不要管网友推荐,我会取消网友推荐的
{an}的前n项和Sn=(3n²+5n)/(n²+3n+2),求lim(n→+∞)[a(n+1)/an]的值等待妙招有刚看到题的不要管网友推荐,我会取消网友推荐的
Sn=(3n^2+9n+6-4n-6)/(n^2+3n+2)=3-2(2n+3)/(n+1)(n+2)=3-2(1/(n+1)+1/(n+2))
关键点S(n-1)=3-2(1/n+1/(n+1))an=Sn-S(n-1)=2(1/n-1/(n+2))a(n+1)=2(1/(n+1)-1/(n+3))a(n+1)/an=(1/(n+1)-1/(n+3))/(1/n-1/(n+2))=n(n+2)/((n+1)(n+3))=(n^2+2n)/(n^2+4n+3)=(1+2/n)/(1+4/n+3/n^2)lima(n+1)/an=1

我要回帖

更多关于 an 1 2 n an 4求通项 的文章

 

随机推荐