若数列{an}中的an=1/(n+1)+1/(n+2)+1/(n+3)+...+1/2n(n∈N*)。
则an+1的表达式是:A:a(n+1)=an+1/(2n+2)B:a(n+1)=an+1/(2n+1)+1/(2n+2)C:a(n+1)=an+1/(2n+1)-1/(2...
则an+1的表达式是:
A:a(n+1)=an+1/(2n+2) B: a(n+1)=an+1/(2n+1)+1/(2n+2)
C:a(n+1)=an+1/(2n+1)-1/(2n+2) D:a(n+1)=an+1/(2n+2)-1/(2n+1)
麻烦过程!谢谢! 展开
A:a(n+1)=an+1/(2n+2) B: a(n+1)=an+1/(2n+1)+1/(2n+2)
C:a(n+1)=an+1/(2n+1)-1/(2n+2) D:a(n+1)=an+1/(2n+2)-1/(2n+1)
麻烦过程!谢谢! 展开
2个回答
展开全部
解:
a(n+1)=1/(n+2)+1/(n+3)+........+1/(2n+2)
=1/(n+2)+1/(n+3)+........+1/2n + 1/(2n+1)+1/(2n+2)
=1/(n+1)+1/(n+2)+1/(n+3)+........+1/2n + 1/(2n+1)+1/(2n+2)-1/(n+1)
=an+ 1/(2n+1)+1/(2n+2)-1/(n+1)
=an+1/(2n+1)-1/(2n+2)
选C
a(n+1)=1/(n+2)+1/(n+3)+........+1/(2n+2)
=1/(n+2)+1/(n+3)+........+1/2n + 1/(2n+1)+1/(2n+2)
=1/(n+1)+1/(n+2)+1/(n+3)+........+1/2n + 1/(2n+1)+1/(2n+2)-1/(n+1)
=an+ 1/(2n+1)+1/(2n+2)-1/(n+1)
=an+1/(2n+1)-1/(2n+2)
选C
追问
请问:an=1/(n+1)+1/(n+2)+1/(n+3)+...+1/2n中1/2n(末项)的前一项是什么?
追答
1/(2n-1)
看前面分母的规律,是连续的自然数。
2012-06-12
展开全部
不知道。
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