$\begin{array}{l}a)\ \text{Đặt $S=\dfrac1{1.2.3}+\dfrac1{2.3.4}+\dfrac1{3.4.5}+\ldots+\dfrac1{14.15.16}$}\\\to 2S=\dfrac2{1.2.3}+\dfrac2{2.3.4}+\dfrac2{3.4.5}+\ldots+\dfrac2{14.15.16}\\\to 2S=\dfrac1{1.2}-\dfrac1{2.3}+\dfrac1{2.3}-\dfrac1{3.4}+\ldots+\dfrac1{14.15}-\dfrac1{15.16}\\\to 2S=\dfrac1{1.2}-\dfrac1{15.16}\\\to2S=\dfrac12-\dfrac1{240}\\\to 2S=\dfrac{119}{240}\\\to S=\dfrac{199}{240}\div2\\\to S=\dfrac{119}{480}\\\,\\b)\ A=\dfrac13+\dfrac1{3^2}+\dfrac1{3^3}+\ldots+\dfrac1{3^8}\\\to 3A=\dfrac33+\dfrac3{3^2}+\dfrac3{3^3}+\ldots+\dfrac3{3^8}\\\to3A=1+\dfrac13+\dfrac1{3^2}+\ldots+\dfrac1{3^7}\\\to3A-A=\left(1+\dfrac13+\dfrac1{3^2}+\ldots+\dfrac1{3^7}\right)-\left(\dfrac13+\dfrac1{3^2}+\dfrac1{3^3}+\ldots+\dfrac1{3^8}\right)\\\to 2A=1-\dfrac1{3^8}\\\to 2A=\dfrac{6560}{6561}\\\to A=\dfrac{6560}{6561}\div2\\\to A=\dfrac{3280}{6561}\\\,\\c)\ \text{- Đặt $B=\dfrac1{2}+\dfrac14+\dfrac18+\ldots+\dfrac1{64}$}\\\to B=\dfrac12+\dfrac1{2^2}+\dfrac1{2^3}+\ldots+\dfrac1{2^6}\\\to 2B=\dfrac22+\dfrac2{2^2}+\dfrac2{2^3}+\ldots+\dfrac2{2^6}\\\to 2B=1+\dfrac12+\dfrac1{2^2}+\ldots+\dfrac1{2^5}\\\to 2B-B=\left(1+\dfrac12+\dfrac1{2^2}+\ldots+\dfrac1{2^5}\right)-\left(\dfrac12+\dfrac1{2^2}+\dfrac1{2^3}+\ldots+\dfrac1{2^6}\right)\\\to B=1-\dfrac1{2^6}\\\to B=\dfrac{63}{64}\\\,\\d)\ \dfrac1{1.2}+\dfrac1{2.3}+\dfrac1{3.4}+\ldots+\dfrac1{99.100}\\=1-\dfrac12+\dfrac12-\dfrac13+\ldots+\dfrac1{99}-\dfrac1{100}\\=1-\dfrac1{100}\\=\dfrac{99}{100}\\\,\\ e)\ \dfrac1{3.5}+\dfrac1{5.7}+\dfrac1{7.9}+\ldots+\dfrac1{99.101}\\=\dfrac12\cdot\left(\dfrac2{3.5}+\dfrac2{5.7}+\dfrac2{7.9}+\ldots+\dfrac2{99.101}\right)\\=\dfrac12\cdot\left(\dfrac13-\dfrac15+\dfrac15-\dfrac17+\ldots+\dfrac1{99}-\dfrac1{101}\right)\\=\dfrac12\cdot\left(\dfrac13-\dfrac1{101}\right)\\=\dfrac12\cdot\dfrac{98}{303}\\=\dfrac{49}{303} \end{array}$
