Prove that \(\frac12+\frac15+\frac18+\frac1{11}+\frac1{20}+\frac1{40}+\frac1{110}+\frac1{1640}=1\).

Solution

1. We observe that

\[ \frac12+\frac15+\frac18+\frac1{11}+\frac1{20}+\frac1{40}+\frac1{110}+\frac1{1640} =\frac12+\frac1{20}+\frac15+\frac18+\frac1{11}+\frac1{110}+\frac1{40}+\frac1{1640} \] \[ =\frac{22}{40}+\frac{13}{40}+\frac{11}{110}+\frac{41}{1640} =\frac{22}{40}+\frac{13}{40}+\frac4{40}+\frac1{40} =\frac{40}{40}=1. \]

Hence, \(\frac12+\frac15+\frac18+\frac1{11}+\frac1{20}+\frac1{40}+\frac1{110}+\frac1{1640}=1\).

2. Find the minimum value of \(A=|x-1|+|x-2|+\cdots+|x-100|\).

Using \(|a|+|b|\geq|a-b|\), we have

\[ |x-k|+|x-(101-k)|\geq|101-2k|,\quad k=1,2,\ldots,50. \]

Therefore,

\[ |x-1|+|x-100|\geq99,\ |x-2|+|x-99|\geq97,\ldots,\ |x-50|+|x-51|\geq1. \]

Adding all inequalities,

\[ A\geq99+97+\cdots+1=\frac{50}{2}(1+99)=2500. \]

The equality holds when \(x\in\bigcap_{k=1}^{50}[k,101-k]=[50,51]\).

Hence,

\[ \boxed{A_{\min}=2500\quad\text{when }x\in[50,51].} \]