作业帮计算:1/[a(a+1)]+1/[(a+1)(a+2)]+1/[(a+2)(a+3)+...+1/[(a+2006)(a+2007)]
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/05 16:11:48
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计算:1/[a(a+1)]+1/[(a+1)(a+2)]+1/[(a+2)(a+3)+...+1/[(a+2006)(a+2007)]
计算:1/[a(a+1)]+1/[(a+1)(a+2)]+1/[(a+2)(a+3)+...+1/[(a+2006)(a+2007)]
计算:1/[a(a+1)]+1/[(a+1)(a+2)]+1/[(a+2)(a+3)+...+1/[(a+2006)(a+2007)]
1/[a(a+1)]=1/a-1/(a+1)
1/[(a+1)(a+2)]=1/(a+1)-1/(a+2)
1/[(a+2)(a+3)]=1/(a+2)-1/(a+3)
……
1/[(a+2005)(a+2006)]=1/(a+2005)-1/(a+2006)
1/[(a+2006)(a+2007)]=1/(a+2006)-1/(a+2007)
原式
1/[a(a+1)]+1/[(a+1)(a+2)]+1/[(a+2)(a+3)+...+1/[(a+2006)(a+2007)]
=1/a-1/(a+1)+1/(a+1)-1/(a+2)+1/(a+2)-1/(a+3)+……+1/(a+2005)-1/(a+2006)+1/(a+2006)-1/(a+2007)
=1/a-1/(a+2007)
=2007/[a(a+2007)]
1/a-1/(a+1)+1/(a+1)-1/(a+2)+......+1/(a+2006)-1/(a+2007)=1/a-1/(a+2007)=2007/a（a+2007） 裂项相销
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