\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+..+100}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+..+\frac{2}{100.101}=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{100}-\frac{1}{101}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{101}\right)=\frac{99}{101}\)