已知|AB-2|与|B-1|互为相反数,试求(1/AB)+[1/(A+1)(B+1)]+[1/(A+2)(B+2)]+...+[1/(A+2008)(B+2008)
因为|AB-2|≥0,|B-1|≥0
已知,|AB-2|与|B-1|互为相反数,则|AB-2|+|B-1|=0
那么,只能是：AB-2=0,且B-1=0
所以,A=2,B=1
那么,原式=(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)+……+(1/2008*2009)
=1-(1/2)+(1/2)-(1/3)+(1/3)-(1/4)+……+(1/2008)-(1/2009)
=1-(1/2009)
=2008/2009