观察下列各式：1/2=1/1*2=1/1-1/2；1/6=1/2*3=1/2-1/3；1/12=1/3*4=1/3-1/4；...
利用得到的结论1/a（a-1）+1/（a-1)(a-2)+...+1/(a-2008)(a-2009)
1/a（a-1）+1/（a-1)(a-2)+...+1/(a-2008)(a-2009)
=1/(a-1)-1/a+1/(a-2)-1/(a-1)+……+1/(a-2009)-1/(a-2008)
=-1/a+1/(a-2009)
=1/(a-2009)-1/a
=2009/a(a-2009)
=1/(a-1)-1/a+1/(a-2)-1/(a-1)+……+1/(a-2009)-1/(a-2008)
=-1/a+1/(a-2009)
=1/(a-2009)-1/a
=2009/a(a-2009)
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第1个回答 2010-06-02
1/a（a-1）+1/（a-1)(a-2)+...+1/(a-2008)(a-2009)
=1/a-1/(a-1)+1/(a-1)-1/(a-2)+...+1/(a-2008)-1/(a-2009)
=1/a-1/(a-2009)
=1/a-1/(a-1)+1/(a-1)-1/(a-2)+...+1/(a-2008)-1/(a-2009)
=1/a-1/(a-2009)
第2个回答 2019-07-13
1、等式右边部分=m*（a-1）+n*（a+2）/(a+2)(a-1)
=（m+n）a+2n-m/(a+2)(a-1)
=4a-1/(a+2)(a-1)
所以2n-m=-1，m+n=4
解得m=3，n=1
2、（1）1/【n(n+1)】=1/n-1/(n+1)
（2）原式=1/(a-2)-1/(a-1)+1/(a-3)-1/(a-2)+1/(a-4)-1/(a-3)+1/(a-5)-1/(a-4)
=1/(a-5)-1/(a-1)
=（m+n）a+2n-m/(a+2)(a-1)
=4a-1/(a+2)(a-1)
所以2n-m=-1，m+n=4
解得m=3，n=1
2、（1）1/【n(n+1)】=1/n-1/(n+1)
（2）原式=1/(a-2)-1/(a-1)+1/(a-3)-1/(a-2)+1/(a-4)-1/(a-3)+1/(a-5)-1/(a-4)
=1/(a-5)-1/(a-1)
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