计算: $\frac{14}{15}+\frac{34}{35}+\frac{62}{63}+\frac{98}{99}+\frac{142}{143}+\frac{194}{195}+\frac{254}{255}+\frac{322}{323}+\frac{398}{399}+\frac{482}{483}$

答案

解：原式 $=10-(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}+\frac{1}{255}+\frac{1}{323}+\frac{1}{399}+\frac{1}{483})$
$=10-(\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+\frac{1}{9\times11}+\frac{1}{11\times13}+\frac{1}{13\times15}+\frac{1}{15\times17}+\frac{1}{17\times19}+\frac{1}{19\times21}+\frac{1}{21\times 23})$
$=10-\frac{1}{2}[(\frac{1}{3}-\frac{1}{5})+(\frac{1}{5}-\frac{1}{7})+(\frac{1}{7}-\frac{1}{9})+(\frac{1}{9}-\frac{1}{11})+(\frac{1}{11}+\frac{1}{13})+(\frac{1}{13}-\frac{1}{15})+(\frac{1}{15}-\frac{1}{17})+(\frac{1}{17}-\frac{1}{19})+(\frac{1}{19}-\frac{1}{21})+(\frac{1}{21}+\frac{1}{23})]$
$=10-\frac{1}{2}(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}+\frac{1}{13}+\frac{1}{13}-\frac{1}{15}+\frac{1}{15}-\frac{1}{17}+\frac{1}{17}-\frac{1}{19}+\frac{1}{19}-\frac{1}{21}+\frac{1}{21}+\frac{1}{23})$
$=10-\frac{1}{2}(\frac{1}{3}-\frac{1}{23})$
$=10-\frac{10}{69}$
$=\frac{680}{69}$