Question

已知|ab-2|与|b-1|互为相反数，求代数式。 1\ab+(a=1)(b+1)\1=(a=2)(b+2)\1+……+(a+2009)(b+2009)\1的值

1\ab+(a+1)(b+1)\1+(a+2)(b+2)\1+……+(a+2009)(b+2009)\1的值

Answer 1

|ab-2|与|b-1|互为相反数,得ab-2=b-1=0，得a=2，b=1；
1/ab+1/(a+1)(b+1)+1/（a+2)(b+2)+……+1/(a+2009)(b+2009)
=1/(1*2)+1/(2*3)+……+1/(2010*2011)
=(2-1)/(1*2)+(3-2)/(2*3)+……+(2011-2010)/(2010*2011)
=2/(1*2)-1(1*2)+……+2011/(2010*2011)-2010/(2010*2011)
=1-1/2+1/2-1/3+……+1/2010-1/2011
=1-1/2011
=2010/2011