1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2012)(b+2012),ab=2,b=1
问题描述：
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2012)(b+2012),ab=2,b=1
错了，是1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2011)(b+2011)+1/（a+2012)（b+2012),ab=2，b=1
答
解由ab=2,b=1知a=2故1/ab=1/1*2=1/1-1/21/(a+1)(b+1)=1/2*3=1/2-1/31/(a+2)(b+2)=1/3*4=1/3-1/4.1/(a+2012)(b+2012)=1/2013*2014=1/2013-1/2014上述各式相加得1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2012)(b+2012...