Question

1/√2+√1=√2-√1;从中找出规律,计算(1/√2+1+1/√3+√2+...+1/√2008+√2007)*(√

Answer 1

(1/√2+1+1/√3+√2+...+1/√2008+√2007)*(√2008+1)

=(√2-1+√3-√2+√4-√3...+√2008-√2007)*(√2008+1)

=[-1+(√2-√2)+(√3-√3)+(√4-√4)+...+(√2007-√2007)+√2008]*(√2008+1)

=[-1+√2008]*(√2008+1)

=(√2008)^2-1^2

=2007