Question

计算1/(1+√2)+1/(√2+√3)+......+1/(√2010+√2011)

这题怎么做,要有过程,√是根号

Answer 1

=1(1-√2)/(1+√2)(1-√2)+1(√2-√3)/(√2+√3)(√2-√3)+.....+1(√2010-√2011)/(√2010+√2011)(√2010-√2011)

=1(1-√2)/(-1)+1(√2-√3)/(-1)+.....+1(√2010-√2011)/(-1)
=(-1+√2)+(-√2+√3)+....+(-√2010+√2011)
=-1+√2011
谢谢采纳，你可以列在草稿纸上看下

Answer 2

1/（√n+√n+1）= （√n-√n+1）/（√n+√n+1）（√n-√n+1）=√n-√n+1
原式=(√2-1)+(√3-√2)+...+(√2011-√2010)=√2011-1

Answer 3

=（1-√2）/（1-2）+（√2-√3）/（2-3）+......+（√2010-√2011）/（2010-2011）
=√2-1+√3-√2+......+√2011-√2010
=√2011-1