1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/(a+2008)(b+2008)= ab=2 b=1 a=2
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1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/(a+2008)(b+2008)= ab=2 b=1 a=2
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/(a+2008)(b+2008)=
ab=2 b=1 a=2
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/(a+2008)(b+2008)= ab=2 b=1 a=2
a=2
b=1
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.1/(a+2008)(b+2008)
=1/1*2+1/2*3+1/3*4+……+1/2009*2010
=1-1/2+1/2-1/3+1/3-1/4+1/4-……-1/2009+1/2009-1/2010
=1-1/2010
=2009/2010
结果为2009/2010。上式=1/2+1/6+1/12+........+1/2010×2009,有前面计算得:1/2+1/6=2/3,2/3+1/12=3/4由此可得:分母应该是a+2008,分子应该是a+2008-1,即2009/2010.
唉20多年都没有算过了,只能看出,它的分母是一个等差数列:
第一项是1/2也是1/(2*1),第二项是1/(3*2)
以此类推,所以数列第N项是1/[(n+1)*n]
那么第(n+1)项是1/{[(n+1)+1]*(n+1)}
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+......1/(a+2008)(b+2008)
=1/1*2+1/2*3+1/3*4+……+1/2009*2010
=1-1/2+1/2-1/3+1/3-1/4+1/4-……-1/2009+1/2009-1/2010
=1-1/2010
=2009/2010