Question

1/ab+1/（a+1)(b+1)+1/(a+2)(b+2)+......1/(a+2006)(b+2006)r如何简便

问题是没有值

Answer 1

1/ab=(b-a)/ab *1/(b-a)=(1/a-1/b)*1/(b-a)
1/(a+1)(b+1)= [(b+1)-(a+1)]/(a+1)(b+1) *1/(b-a)=[1/(a+1)-1/(b+1)]*1/(b-a)
....
1/(a+2006)(b+2006)=[(b+2006)-(a+2006)]/(a+2006)(b+2006) *1/(b-a)=[1/(a+2006)-1/(b+2006)]*1/(b-a)

原式=1/(b-a)* {(1/a-1/b)+[1/(a+1)-1/(b+1)]+....+[1/(a+2006)-1/(b+2006)]}
=1/(b-a)*{[1/a+1/(a+1)+...+1/(a+2006)]-[1/b+1/(b+1)+....+1/(b+2006)]}

只能做到这一步,如果还要化简,当知道a,b的差值后,可以消去很多项.