{n(n+1)(n+2)}分之一怎样求和
2个回答
展开全部
1/(n(n+1)(n+2)
=(1/2)*2/n(n+1)(n+2)
=(1/2)*[(n+2)-n]/n(n+1)(n+2)
=(1/2)*[(n+2)/n(n+1)(n+2)-n/n(n+1)(n+2)]
=(1/2)*[1/n(n+1)-1/(n+1)(n+2)]
所以1/1*2*3+……+1/n(n+1)(n+2)
=(1/2)*[1/1*2-1/2*3+……+1/n(n+1)-1/(n+1)(n+2)]
=(1/2)*[1/1*2-1/(n+1)(n+2)]
=(n²+3n)/(4n²+12n+8)
=(1/2)*2/n(n+1)(n+2)
=(1/2)*[(n+2)-n]/n(n+1)(n+2)
=(1/2)*[(n+2)/n(n+1)(n+2)-n/n(n+1)(n+2)]
=(1/2)*[1/n(n+1)-1/(n+1)(n+2)]
所以1/1*2*3+……+1/n(n+1)(n+2)
=(1/2)*[1/1*2-1/2*3+……+1/n(n+1)-1/(n+1)(n+2)]
=(1/2)*[1/1*2-1/(n+1)(n+2)]
=(n²+3n)/(4n²+12n+8)
本回答被提问者采纳
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
