已知a−1+（ab-2）2=0，求1ab+1(a+1)(b+1)+1(a+2)(b+2)+…+1(a+2005)(b+2005)的值．

Question

已知√(a−1)+（ab-2）²=0，求(1)/(ab)+(1)/((a+1)(b+1))+(1)/((a+2)(b+2))+…+(1)/((a+2005)(b+2005))的值．

Answer 1

∵√(a−1)+（ab-2）²=0，
∴a-1=0且ab-2=0，
解得：a=1，b=2，
∴(1)/(ab)+(1)/((a+1)(b+1))+…+(1)/((a+2005)(b+2005))
=(1)/(2)+(1)/(2×3)+(1)/(3×4)+…+(1)/(2006×2007)
=(1)/(2)+(1)/(2)-(1)/(3)+(1)/(3)-(1)/(4)+…+(1)/(2006)-(1)/(2007)
=(1)/(2)+(1)/(2)-(1)/(2007)
=(2006)/(2007)．