Ta có :
\(100-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)
\(=100-\left[1+\left(1-\frac{1}{2}\right)+\left(1-\frac{2}{3}\right)+...+\left(1-\frac{99}{100}\right)\right]\)
\(=100-\left[\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\right]\)
\(=100-\left[100-\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\right]\)
\(=100-100+\left(\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\right)\)
\(=\frac{1}{2}+\frac{2}{3}+...+\frac{99}{100}\)
ta có 100-(1+1/2+1/3+.....+1/100)
=(1+1+1......1)(99 số 1)-(1+1/2+1/3+......+1/100)
=(1-1)+(1-1/2)+(1-1/3)+.......+(1-1/100)
=1/2+2/3+3/4+.....+99/100
Ta có:
100-(1+1/2+1/3+...+1/100)
=(1-1)+(1-1/2)+(1-1/3)+...+(1-1/100)
=1/2+2/3+...+99/100
