计算:{1/a(a+1)}+{1/(a+1)(a+2)}+{1/(a+2)(a+3)}+…+{1/(a+2006)(a+
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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)}+{1/(a+2007(a+2008)}
这个,我们来分析一下,
首先,这是一个有规律可以找的,
一,分母是两个数的积,并且,相差1
二,分子,都是 1
三,这样,想到,如果,分子,分母相差1,且是积,
如1/ab,a-b=1,这时,1/ab=1/b-1/a
四,做,:
{1/a(a+1)}+{1/(a+1)(a+2)}+{1/(a+2)(a+3)}+…+{1/(a+2006)(a+2007)}+{1/(a+2007(a+2008)}
=1/a-1/(a+1)+1/(a+1)-1/(a+2)+1/(a+2)-1/(a+3)+…+1/(a+2006)-1/(a+2007)+1/(a+2007)-1/(a+2008)
=1/a-1/(a+2008)
=2008/a(a+2008)
首先,这是一个有规律可以找的,
一,分母是两个数的积,并且,相差1
二,分子,都是 1
三,这样,想到,如果,分子,分母相差1,且是积,
如1/ab,a-b=1,这时,1/ab=1/b-1/a
四,做,:
{1/a(a+1)}+{1/(a+1)(a+2)}+{1/(a+2)(a+3)}+…+{1/(a+2006)(a+2007)}+{1/(a+2007(a+2008)}
=1/a-1/(a+1)+1/(a+1)-1/(a+2)+1/(a+2)-1/(a+3)+…+1/(a+2006)-1/(a+2007)+1/(a+2007)-1/(a+2008)
=1/a-1/(a+2008)
=2008/a(a+2008)
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