Question

计算1/（2√1+√2）+1/（3√2+2√3）+……+1/（100√99+99√100）,

Answer 1

容易看出每项的分母都是 (n+1)√n + n√(n+1) 的形式,所以先看一个一般的情况.即先来计算 1/[(n+1)√n+n√(n+1)].
1/[(n+1)√n+n√(n+1)] (分子分母同时乘以 (n+1)√n - n√(n+1),即分母有理化)
=[(n+1)√n-n√(n+1)]/[n(n+1)^2-(n+1)n^2] (化简分母)
=[(n+1)√n-n√(n+1)]/[n(n+1)]
=(n+1)√n/[n(n+1)] - n√(n+1)/[n(n+1)]
=√n/n - √(n+1)/(n+1)
=1/√n - 1/√(n+1)
于是
1/(2√1+√2)+1/(3√2+√3)+...+1/(100√99+99√100)
=(1/√1-1/√2)+(1/√2-1/√3)+...+(1/√99-1/√100)
=1/√1 - 1/√100
=1-1/10
=9/10
即原式 = 9/10.