设数列{an}满足a1-2,a(n+1)=n/(n+2)*an(n∈N*).
(1).求数列{an}的通项公式；(2).求数列{an/4}的前2011项和S做第二题就好了...
(1).求数列{an}的通项公式； (2).求数列{an/4}的前2011项和S
做第二题就好了 展开
做第二题就好了 展开
1个回答
展开全部
解：（1）
a(n+1)=n/(n+2)*an=n/(n+2)*(n-1)/(n+1)a(n-1)=n/(n+2)*(n-1)/(n+1)*(n-2)/n*a(n-2)
=n/(n+2)*(n-1)/(n+1) * (n-2)/(n) * (n-3)/(n-1)*......*3/5*2/4*1/3*a1
=2/(n+1)(n+2)*a1
=4/(n+1)(n+2)
an=4/n(n+1) (n∈N*)
(2){an/4}={1/n(n+1)}
1/n(n+1)=1/n-1/(n+1)
S=(1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/2011-1/2012)
=1-1/2012
=2011/2012
a(n+1)=n/(n+2)*an=n/(n+2)*(n-1)/(n+1)a(n-1)=n/(n+2)*(n-1)/(n+1)*(n-2)/n*a(n-2)
=n/(n+2)*(n-1)/(n+1) * (n-2)/(n) * (n-3)/(n-1)*......*3/5*2/4*1/3*a1
=2/(n+1)(n+2)*a1
=4/(n+1)(n+2)
an=4/n(n+1) (n∈N*)
(2){an/4}={1/n(n+1)}
1/n(n+1)=1/n-1/(n+1)
S=(1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/2011-1/2012)
=1-1/2012
=2011/2012
本回答由提问者推荐
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
