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

问题描述:

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

(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)=(...