Question

已知a-1的平方根加(ab-2)的平方=0求ab分之1加(a+1)(b+1)分之1加(a+2)(b+2)分之1加……(a+2008)(b+2008)...

已知a-1的平方根加(ab-2)的平方=0求ab分之1加(a+1)(b+1)分之1加(a+2)(b+2)分之1加……(a+2008)(b+2008)分之1的值

Answer 1

已知a-1的平方根加(ab-2)的平方=0
a-1=0 a=1
ab-2=0 b=2
ab分之1加(a+1)(b+1)分之1加(a+2)(b+2)分之1加……(a+2008)(b+2008)分之1

=1/1x2+1/2x3+1/3x4+……+1/2009x2010
=1-1/2+1/2-1/3+1/3-1/4+……+1/2009-1/2010
=1-1/2010
=2009/2010

Answer 2

√(a-1)+(ab-2)²=0
=>a=1,b=2
=>1/ab+1/(a+1)(b+1)+....+1/(a+2008)(b+2008)
=1-1/2+1/2-1/3+1/3-1/4+...+1/2008-1/2009+1/2009-1/2010
=1-1/2010
=2009/2010